NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Ans : (i) Similar
(ii) Similar 
(iii) Equilateral
(iv) (a) Equal
(b) Proportional 

Ans : (i) Two equilateral triangles with sides 1 cm and 2 cm.

Two squares with sides 1 cm and 2 cm.

(ii) Trapezium and square
 

Ans : Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e 1:2, but their corresponding angles are not equal. 

Ans : (i) 

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\( \begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{1.5}{3}=\frac{1}{x}} \\ {x=\frac{3 \times 1}{1.5}} \\ {x=2} \\ {\therefore \mathrm{EC}=2 \mathrm{cm}}\end{array} \) 
(ii) 

Let AD = x cm 
It is given that DE || BC
By using basic proportionality theorem, we obtain
\( \begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{x}{7.2}=\frac{1.8}{5.4}} \\ {x=\frac{1.8 \times 7.2}{5.4}} \\ {x=2.4} \\ {\therefore \mathrm{AD}=2.4 \mathrm{cm}}\end{array} \) 

Ans : (i) 

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\( \frac{P E}{E Q}=\frac{3.9}{3}=1.3 \) 
\( \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3.6}{2.4}=1.5 \) 
Hence, \( \frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}} \) 
Therefore, EF is not parallel to QR.
(ii)  

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
\( \frac{P E}{E Q}=\frac{4}{4.5}=\frac{8}{9} \) 
\( \frac{P F}{F R}=\frac{8}{9} \) 
Hence \( \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}} \) 
\( Therefore, EF is parallel to QR. \) 
(iii)  

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
\( \frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64} \) 
\( \frac{\mathrm{PF}}{\mathrm{PR}}=\frac{0.36}{2.56}=\frac{9}{64} \) 
\( \begin{array}{l}{\text { Hence, } \frac{\mathrm{PE}}{\mathrm{PQ}}=\frac{\mathrm{PF}}{\mathrm{PR}}} \\ {\text { Therefore, EF is parallel to QR. }}\end{array} \) 

Ans : 
In the given figure, LM || CB
By using basic proportionality theorem, we obtain
\( \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AL}}{\mathrm{AC}} \) ..(i)
Similarly, LN || CD
\( \therefore \frac{\mathrm{AN}}{\mathrm{AD}}=\frac{\mathrm{AL}}{\mathrm{AC}} \) 
(ii)
From (i) and (ii), we obtain
\( \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}} \) 

Ans : 
In \( \triangle A B C, D E\|A C \) 
\( \therefore \frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BE}}{\mathrm{EC}} \) (Basic Proportionality Theorem) (i) 

In \( \triangle B A E, D F\|A E \) 
\( \therefore \frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BF}}{\mathrm{FE}} \quad(\text { Basic Proportionality Theorem) } \) (ii) 
From (i) and (ii), we obtain
\( \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}} \) 

Ans : 
In \( \triangle \mathrm{POQ}, \mathrm{DE}\|\mathrm{OQ} \) 
\( \therefore \frac{P E}{E Q}=\frac{P D}{D O} \quad(\text { Basic proportionality theorem }) \) (i) 

In \( \triangle \mathrm{POR}, \mathrm{DF}\|\mathrm{OR} \) 
\( \therefore \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{\mathrm{PD}}{\mathrm{DO}} \quad \) (Basic proportionality theorem) 
(ii) 
From (i) and (ii), we obtain
\( \frac{P E}{E Q}=\frac{P F}{F R} \) 
\( \therefore \mathrm{EF}\|\mathrm{OR} \quad \text { (Converse of basic proportionality theorem) } \) 
 

Ans : 
In \( \triangle \mathrm{POQ}, \mathrm{AB}\|\mathrm{PQ} \) 
\( \therefore \frac{\mathrm{O} \mathrm{A}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}} \quad \) (Basic proportionality theorem) (i) 

In \( \triangle \mathrm{POR}, \mathrm{AC}\|\mathrm{PR} \) 
\( \therefore \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}} \quad \) (By basic proportionality theorem) (ii) 
From (i) and (ii), we obtain.
\( \frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}} \) 
\( \therefore \mathrm{BC}\|\mathrm{QR} \) (By the converse of basic proportionality theorem)
 

Ans : 
Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ || BC
By using basic proportionality theorem, we obtain
\( \frac{\mathrm{AQ}}{\mathrm{QC}}=\frac{\mathrm{AP}}{\mathrm{PB}} \) 
\( \frac{\mathrm{AQ}}{\mathrm{QC}}=\frac{1}{1} \quad(\mathrm{P} \text { is the mid-point of } \mathrm{AB} . \therefore \mathrm{AP}=\mathrm{PB}) \) 
\( \Rightarrow A Q=Q C \) 
Or, Q is the mid-point of AC. 

Ans : 
Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.
I.e., AP = PB and AQ = QC
It can be observed that
\( \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{1}{1} \) 
and \( \frac{\mathrm{AQ}}{\mathrm{QC}}=\frac{1}{1} \) 
\( \therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}} \) 
Hence, by using basic proportionality theorem, we obtain
\( \mathrm{PQ}\|\mathrm{BC} \) 

Ans : 
Draw a line EF through point O, such that EF || CD
In \( \triangle A D C, E O\|C D \) 
By using basic proportionality theorem, we obtain
\( \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}} \) ...(1)
In \( \triangle \mathrm{ABD}, \text { OE }\|\mathrm{AB} \) 
So, by using basic proportionality theorem, we obtain
\( \frac{\mathrm{ED}}{\mathrm{AE}}=\frac{\mathrm{OD}}{\mathrm{BO}} \) 
\( \Rightarrow \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{OD}} \)... (2)
From equations (1) and (2), we obtain
\( \begin{array}{l}{\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{OD}}} \\ {\Rightarrow \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OD}}}\end{array} \) 

Ans : Let us consider the following figure for the given question.

Draw a line OE || AB

In \( \triangle A B D, O E\|A B \) 
By using basic proportionality theorem, we obtain
\( \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{OD}} \) (1)
However, it is given that 
\( \frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}} \) (2)
From equations (1) and (2), we obtain
\( \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}} \) 
\( \Rightarrow \text { EO }\|\text { DC }[\mathrm{By} \text { the converse of basic proportionality theorem] } \) 
\( \Rightarrow A B\|O E\| D C \) 
\( \begin{array}{l}{\Rightarrow A B\|C D} \\ {\therefore A B C D \text { is a trapezium. }}\end{array} \) 

Ans : (i) \( \angle A=\angle P=60^{\circ} \) 
\( \angle B=\angle Q=80^{\circ} \) 
\( \angle C=\angle R=40^{\circ} \) 
Therefore, \( \triangle \mathrm{ABC}-\triangle \mathrm{PQR}[\mathrm{By} \text { AAA similarity criterion }] \) 
\( \frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{CA}}{\mathrm{PQ}} \) 

(ii) \( \therefore \triangle \mathrm{ABC}-\Delta \mathrm{QRP} \quad[\text { By SSS similarity criterion }] \) 

(iii) The given triangles are not similar as the corresponding sides are not proportional.

(iv) The given triangles are not similar as the corresponding sides are not proportional.

(v) THe given triangles are not similar as the corresponding sides are not proportional.

(vi) In \( \Delta \mathrm{DEF} \) 
\( \angle D+\angle E+\angle F=180^{\circ} \) 
( Sum of the measures of the angles of a triangle is \( 180^{\circ} \). )
\( 70^{\circ}+80^{\circ}+\angle F=180^{\circ} \) 
\( \angle F=30^{\circ} \) 
Similarly, In \( \triangle \mathrm{PQR} \) 
\( \angle P+\angle Q+\angle R=180^{\circ} \) 
( Sum of the measures of the angles of a triangle is \( 180^{\circ} \). )
\( \begin{array}{l}{\angle P+80^{\circ}+30^{\circ}=180^{\circ}} \\ {\angle P=70^{\circ}}\end{array} \) 
In \( \triangle D E F \text { and } \Delta P Q R, \) 
\( \angle D=\angle P\left(E a c h 70^{\circ}\right) \) 
\( \angle E=\angle Q\left(E a c h 80^{\circ}\right) \) 
\( \angle F=\angle R\left(\text { Each } 30^{\circ}\right) \) 
\( \therefore \triangle \mathrm{DEF} \sim \triangle \mathrm{PQR}[\mathrm{By} \text { AAA similarity criterion }] \) 

Ans : DOB is a straight line.
\( \therefore \angle D O C+\angle C O B=180^{\circ} \) 
\( \begin{array}{l}{\Rightarrow \angle D O C=180^{\circ}-125^{\circ}} \\ {=55^{\circ}}\end{array} \) 
In \( \Delta D O C \) 
\( \angle D C O+\angle C D O+\angle D O C=180^{\circ} \) 
( Sum of the measures of the angles of a triangle is \( 180^{\circ} \). )
\( \begin{array}{l}{\Rightarrow \angle D C O+70^{\circ}+55^{\circ}=180^{\circ}} \\ {\Rightarrow \angle D C O=55^{\circ}}\end{array} \) 
It is given that \( \triangle O D C-\triangle O B A \) 
\( \therefore \angle O A B=\angle O C D[\text { Corresponding angles are equal in similar triangles.] } \) 
\( \Rightarrow \angle O A B=55^{\circ} \) 

Ans : 
In \( \triangle D O C \text { and } \triangle B O A \),
\( \angle \mathrm{CDO}=\angle \mathrm{ABO}[\text { Alternate interior angles as } \mathrm{AB} \| \mathrm{CD}] \) 
\( \angle \mathrm{DCO}=\angle \mathrm{BAO}[\text { Alternate interior angles as } \mathrm{AB} \| \mathrm{CD}] \) 
\( \angle D O C=\angle B O A[\text { Vertically opposite angles }] \) 
\( \therefore \triangle D O C-\triangle B O A \) [AAA similarity criterion]
\( \therefore \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}} \) [Corresponding sides are proportional]
\( \Rightarrow \frac{\mathrm{O} \mathrm{A}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}} \) 

Ans : 
In \( \Delta P Q R, \angle P Q R=\angle P R Q \) 
\( \therefore \mathrm{PQ}=\mathrm{PR}(\mathrm{i}) \) 
Given, 
\( \frac{\mathrm{Q} \mathrm{R}}{\mathrm{OS}}=\frac{\mathrm{QT}}{\mathrm{PR}} \) 
Using (i), we obtain
\( \frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}} \) (ii) 
In \( \triangle \mathrm{PQS} \text { and } \Delta \mathrm{TQR} \) 
\( \frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}} \quad[\mathrm{Using}(i i)] \) 
\( \angle Q=\angle Q \) 
\( \therefore \triangle \mathrm{PQS} \sim \triangle \mathrm{TQR} \) [SAS similarity criterion] 

Ans : 
In \( \triangle \mathrm{RPQ} \text { and } \Delta \mathrm{RST} \), 
\( \angle \mathrm{RTS}=\angle \mathrm{QPS}(\text { Given }) \) 
\( \begin{array}{l}{\angle R=\angle R(\text { Common angle })} \\ {\therefore \triangle R P Q-\Delta R T S(B y \text { AA similarity criterion) }}\end{array} \) 

Ans : It is given that \( \triangle \mathrm{ABE} \cong \triangle \mathrm{ACD} \) 
\( \therefore A B=A C[B y C P C T](1) \) 
\( \begin{array}{l}{\text { And, } A D=A E[B Y C P C T](2)} \\ {\text { In } \triangle A D E \text { and } \triangle A B C}\end{array} \) 
\( \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}[\text { Dividing equation }(2) \mathrm{by}(1)] \) 
\( \angle A=\angle A[\text { Common angle }] \) 
\( \therefore \triangle \mathrm{ADE}-\triangle \mathrm{ABC}[\mathrm{By} \text { SAS similarity criterion] } \) 

Ans : (i) 

In \( \triangle A E P \text { and } \triangle C D P. \) 
\( \angle A E P=\angle C D P\left(E a c h 90^{\circ}\right) \) 
\( \angle \mathrm{APE}=\angle \mathrm{CPD}(\text { Vertically opposite angles }) \) 
Hence, by using AA similarity criterion,
\( \triangle \mathrm{AEP} \sim \Delta \mathrm{CDP} \) 

(ii) 

In \( \triangle \mathrm{ABD} \text { and } \Delta \mathrm{CBE} \) 
\( \angle A D B=\angle C E B\left(E a c h 90^{\circ}\right) \) 
\( \angle A B D=\angle C B E(\text { Common }) \) 
\( \begin{array}{l}{\text { Hence, by using AA similarity criterion, }} \\ {\triangle A B D-\Delta C B E}\end{array} \) 

(iii) 

In \( \triangle \mathrm{AEP} \text { and } \triangle \mathrm{ADB} \) ,
\( \begin{aligned} \angle A E P &=\angle A D B\left(E a c h 90^{\circ}\right) \\ \angle P A E &=\angle D A B(\text { Common }) \end{aligned} \) 
Hence, y using AA similarity criterion,
\( \triangle \mathrm{AEP} \sim \triangle \mathrm{ADB} \) 
(iv) 
In \( \Delta \mathrm{PDC} \text { and } \Delta \mathrm{BEC} \) 
\( \begin{aligned} \angle \mathrm{PDC} &=\angle \mathrm{BEC}\left(\text { Each } 90^{\circ}\right) \\ \angle \mathrm{PCD} &=\angle \mathrm{BCE}(\text { Common angle }) \end{aligned} \) 
Hence, by using AA similarity criterion,
\( \triangle P D C-\triangle B E C \) 

Ans : 
In \( \triangle \mathrm{ABE} \text { and } \Delta \mathrm{CFB} \),
\( \angle \mathrm{A}=\angle \mathrm{C}(\text { Opposite angles of a parallelogram) } \) 
\( \angle A E B=\angle C B F(\text { Alternate interior angles as } A E \| B C) \) 
\( \therefore \triangle \mathrm{ABE}-\Delta \mathrm{CFB}(\mathrm{By} \text { AA similarity criterion) } \) 

Ans : In \( \triangle A B C \text { and } \triangle A M P, \) 
\( \angle A B C=\angle A M P\left(E a c h 90^{\circ}\right) \) 
\( \angle A=\angle A(\text { Common }) \) 
\( \therefore \triangle \mathrm{ABC}-\triangle \mathrm{AMP}(\mathrm{By} \text { AA similarity criterion) } \) 
\( \Rightarrow \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}} \quad \) ( Corresponding sides of similar triangles are proportional) 

Ans : 
It is given that \( \triangle \mathrm{ABC}-\Delta \mathrm{FEG} \) 
\( \therefore \angle A=\angle F, \angle B=\angle E, \text { and } \angle A C B=\angle F G E \) 
\(\begin{array}{l}{\angle A C B=\angle F G E} \\ {\therefore \angle A C D=\angle F G H(\text { Angle bisector })}\end{array} \) 
\( \begin{array}{l}{\text { And, } \angle \mathrm{DCB}=\angle \mathrm{HGE} \text { (Angle bisector) }} \\ {\text { In } \triangle \mathrm{ACD} \text { and } \Delta \mathrm{FGH} \text { , }}\end{array} \) 
\( \begin{array}{l}{\angle \mathrm{A}=\angle \mathrm{F}(\text { Proved above) }} \\ {\angle \mathrm{ACD}=\angle \mathrm{FGH}(\text { Proved above })}\end{array} \) 
\( \therefore \triangle A C D \sim \Delta F G H(B y \text { AA similarity criterion) } \) 
\( \Rightarrow \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}} \) 
In \( \triangle D C B \text { and } \Delta H G E, \) 
\( \begin{array}{l}{\angle \mathrm{DCB}=\angle \mathrm{HGE}(\text { Proved above })} \\ {\angle \mathrm{B}=\angle \mathrm{E}(\text { Proved above })}\end{array} \) 
\( \triangle \triangle D C B \sim \Delta H G E(B y \text { AA similarity criterion) } \) 
\( \begin{array}{l}{\text { In } \triangle D C A \text { and } \Delta H G F} \\ {\angle A C D=\angle F G H(\text { Proved above })}\end{array} \) 
\( \angle A=\angle F(\text { Proved above }) \) 
\( \therefore \triangle \mathrm{DCA} \sim \Delta \mathrm{HGF}(\text { By AA similarity criterion) } \) 

Ans : It is given that ABC is an isosceles triangle.
\( \begin{array}{l}{\therefore \mathrm{AB}=\mathrm{AC}} \\ {\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{ECF}} \\ {\text { In } \triangle \mathrm{ABD} \text { and } \Delta \mathrm{ECF}}\end{array} \)
\( \angle A D B=\angle E F C\left(E a c h 90^{\circ}\right) \) 
\( \angle B A D=\angle C E F(\text { Proved above }) \) 
\( \therefore \triangle \mathrm{ABD} \sim \Delta \mathrm{ECF}(\mathrm{By} \text { using } \mathrm{AA} \text { similarity criterion }) \) 

Ans : Median divides the opposite side.
\( \therefore \) \( \mathrm{BD}=\frac{\mathrm{BC}}{2} \text { and } \mathrm{QM}=\frac{\mathrm{QR}}{2} \) 
\( \begin{array}{l}{\text { Given that, }} \\ {\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}}\end{array} \) 
\( \begin{array}{l}{\Rightarrow \frac{A B}{P Q}=\frac{\frac{1}{2} B C}{\frac{1}{2} Q R}=\frac{A D}{P M}} \\ {\Rightarrow \frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}}\end{array} \) 
In \( \triangle A B D \text { and } \Delta P Q M \) 
\( \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \) (Proved above)
\( \triangle \triangle \mathrm{ABD} \sim \triangle \mathrm{PQM}(\mathrm{By} \text { SSS similarity criterion }) \) 
\( \Rightarrow \angle A B D=\angle P Q M(\text { Corresponding angles of similar triangles) } \) 
\( \begin{array}{l}{\text { In } \triangle A B C \text { and } \Delta P Q R \text { , }} \\ {\angle A B D=\angle P Q M \text { (Proved above) }}\end{array} \) 
\( \begin{array}{l}{\frac{A B}{P Q}=\frac{B C}{Q R}} \\ {\therefore \triangle A B C \sim \Delta P Q R(B y \text { SAS similarity criterion) }}\end{array} \) 

Ans : 
In \( \triangle A D C \text { and } \Delta B A C \),
\( \begin{aligned} \angle A D C &=\angle B A C(\text { Given }) \\ \angle A C D &=\angle B C A(\text { Common angle }) \end{aligned} \) 
\( \therefore \triangle A D C \sim \Delta B A C(B y \text { AA similarity criterion) } \) 
We know that corresponding sides of similar triangles are in proportion.
\( \therefore \frac{\mathrm{CA}}{\mathrm{CB}}=\frac{\mathrm{CD}}{\mathrm{CA}} \) 
\( \Rightarrow \mathrm{CA}^{2}=\mathrm{CB} \times \mathrm{CD} \) 

Ans : 
Given that,
\( \frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M} \) 
Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.
 
We know that medians divide opposite sides.
Therefore, BD = DC and QM = MR 
Also, AD = DE (By construction)
And, PM = ML(By construction)
In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
\( \therefore A C=B E \text { and } A B=E C \) (Opposite sides of a parallelogram are equal)
Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR
It was given that 
\( \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}} \) 
\( \Rightarrow \frac{A B}{P Q}=\frac{B E}{Q L}=\frac{2 A D}{2 P M} \) 
\( \Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QL}}=\frac{\mathrm{AE}}{\mathrm{PL}} \) 
\( \therefore \triangle \mathrm{ABE} \sim \triangle \mathrm{PQL} \) (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
\( \therefore \angle B A E=\angle Q P L \ldots(1) \) 
Similarly, it can be proved that \( \triangle \mathrm{AEC} \sim \triangle \mathrm{PLR} \) and \( \angle C A E=\angle R P L \ldots(2) \) 
Adding equation (1) and (2), we obtain
\( \angle B A E+\angle C A E=\angle Q P L+\angle R P L \) 
\( \Rightarrow \angle C A B=\angle R P Q \ldots(3) \) 
In \( \triangle A B C \text { and } \Delta P Q R_{\prime} \)
\( \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}} \) (Given)
\( \angle C A B=\angle R P Q[U \text { sing equation }(3)] \) 
\( \therefore \triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}(\mathrm{By} \text { SAS similarity criterion) } \) 

Ans : 
Let AB and CD be a tower and a pole respectively.
Let the shadow of BE and DF be the shadow of AB and CD respectively.
At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.
Therefore, \( \angle D C F=\angle B A E \) 
And, \( \angle \mathrm{DFC}=\angle \mathrm{BEA} \) 
\( \angle \mathrm{CDF}=\angle \mathrm{ABE}(\text { Tower and pole are vertical to the ground) } \) 
\( \therefore \triangle \mathrm{ABE} \sim \triangle \mathrm{CDF} \) (AAA similarity criterion)
\( \Rightarrow \frac{\mathrm{AB}}{\mathrm{CD}}=\frac{\mathrm{BE}}{\mathrm{DF}} \) 
\( \Rightarrow \frac{\mathrm{AB}}{6 \mathrm{m}}=\frac{28}{4} \) 
\( \Rightarrow A B=42 \mathrm{m} \) 
Therefore, the height of the tower will be 42 metres. 

Ans : 
It is given that \( \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \) 
We know that the corresponding sides of similar triangles are in proportion.
\( \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}} \) …(1)
Since AD and PM are medians, they will divide their opposite sides.
\( \mathrm{BD}=\frac{\mathrm{BC}}{2} \text { and } \mathrm{QM}=\frac{\mathrm{QR}}{2} \) …(3)
From equations (1) and (3), we obtain
\( \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}} \) …(4)
In \( \triangle \mathrm{ABD} \text { and } \Delta \mathrm{PQM} \) 
\( \angle B=\angle Q[U \sin g \text { equation }(2)] \) 
\( \frac{A B}{P Q}=\frac{B D}{Q M} \) [Using equation (4)]
\( \therefore \triangle \mathrm{ABD} \sim \Delta \mathrm{PQM} \) (By SAS similarity criterion)
\( \Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \) 

Ans : It is given that \( \triangle \mathrm{ABC}-\Delta \mathrm{DEF} \) .
\( \therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{\mathrm{AB}}{\mathrm{DE}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{DF}}\right)^{2} \) 
Given that,
\( \mathrm{EF}=15.4 \mathrm{cm} \),
\( \begin{aligned} \operatorname{ar}(\Delta \mathrm{ABC}) &=64 \mathrm{cm}^{2} \\ \operatorname{ar}(\Delta \mathrm{DEF}) &=121 \mathrm{cm}^{2} \end{aligned} \) 
\( \therefore \frac{\operatorname{ar}(\mathrm{ABC})}{\operatorname{ar}(\mathrm{DEF})}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)^{2} \) 
\( \Rightarrow\left(\frac{64 \mathrm{cm}^{2}}{121 \mathrm{cm}^{2}}\right)=\frac{\mathrm{BC}^{2}}{(15.4 \mathrm{cm})^{2}} \) 
\( \Rightarrow \frac{\mathrm{BC}}{15.4}=\left(\frac{8}{11}\right) \mathrm{cm} \) 
\( \Rightarrow \mathrm{BC}=\left(\frac{8 \times 15.4}{11}\right) \mathrm{cm}=(8 \times 1.4) \mathrm{cm}=11.2 \mathrm{cm} \) 

Ans : 
Since AB || CD,
\( \therefore \angle O A B=\angle O C D \text { and } \angle O B A=\angle O D C(\text { Alternate interior angles) } \) 
In \( \triangle \mathrm{AOB} \text { and } \Delta \mathrm{COD} , \)
\( \angle \mathrm{AOB}=\angle \mathrm{COD} \) (Vertically opposite angles)
\( \angle O A B=\angle O C D(\text { Alternate interior angles }) \) 
\( \angle O B A=\angle O D C(\text { Alternate interior angles }) \) 
\( \therefore \triangle \mathrm{AOB} \sim \Delta \mathrm{COD}(\mathrm{By} \text { AAA similarity criterion }) \) 
\( \therefore \frac{\operatorname{ar}(\Delta \mathrm{AOB})}{\operatorname{ar}(\Delta \mathrm{COD})}=\left(\frac{\mathrm{AB}}{\mathrm{CD}}\right)^{2} \) 
Since \( A B=2 C D \) 
\( \therefore \frac{\operatorname{ar}(\Delta \mathrm{AOB})}{\operatorname{ar}(\Delta \mathrm{COD})}=\left(\frac{2 \mathrm{CD}}{\mathrm{CD}}\right)^{2}=\frac{4}{1}=4 : 1 \) 

Ans : Let us draw two perpendicular AP and DM on line BC.

We know that area of a triangle = 1/2 x Base x Height
\( \therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DBC})}=\frac{\frac{1}{2} \mathrm{BC} \times \mathrm{AP}}{\frac{1}{2} \mathrm{BC} \times \mathrm{DM}}=\frac{\mathrm{AP}}{\mathrm{DM}} \) 
In \( \triangle \mathrm{APO} \text { and } \triangle \mathrm{DMO} \) 
\( \angle A P O=\angle D M O\left(E a c h=90^{\circ}\right) \) 
\( \angle \mathrm{AOP}=\angle \mathrm{DOM}(\text { Vertically opposite angles) } \) 
\( \therefore \triangle \mathrm{APO} \sim \triangle \mathrm{DMO}(\mathrm{By} \text { AA similarity criterion) } \) 
\( \therefore \frac{A P}{D M}=\frac{A O}{D O} \) 
\( \Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}} \) 

Ans : Let us assume two similar triangles are \( \triangle \mathrm{ABC} \sim \Delta \mathrm{PQR} \)
\( \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)^{2} \) (1)
Given that, ar \( (\Delta \mathrm{ABC})=\text { ar }(\Delta \mathrm{PQR}) \) 
\( \Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=1 \) 
Putting this value in equation (1), we obtain
\( \mathrm{I}=\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)^{2} \) 
\( \Rightarrow \mathrm{AB}=\mathrm{PQ}, \mathrm{BC}=\mathrm{QR}, \text { and } \mathrm{AC}=\mathrm{PR} \) 
\( \therefore \Delta \mathrm{ABC} \cong \Delta \mathrm{PQR} \quad(\text { By SSS congruence criterion }) \) 

Ans : 
\( \therefore \mathrm{DE}\left\|\mathrm{AC} \text { and } \mathrm{DE}=\frac{1}{2} \mathrm{AC}\right. \) 
In \( \triangle B E D \text { and } \Delta B C A, \) 
\( \angle B E D=\angle B C A \) (Corresponding angles)
\( \angle \mathrm{BDE}=\angle \mathrm{BAC} \) (Corresponding angles)
\( \angle \mathrm{EBD}=\angle \mathrm{CBA} \) ( Common angles)
\( \therefore \triangle B E D \sim \Delta B C A \) (AAA similarity criterion)
\( \Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{BED})}{\operatorname{ar}(\Delta \mathrm{BCA})}=\frac{1}{4} \) 
\( \Rightarrow \operatorname{ar}(\Delta \mathrm{BED})=\frac{1}{4} \operatorname{ar}(\Delta \mathrm{BCA}) \) 
Similarly, \( \operatorname{ar}(\Delta \mathrm{CFE})=\frac{1}{4} \operatorname{ar}(\mathrm{CBA}) \text { and ar }(\Delta \mathrm{ADF})=\frac{1}{4} \operatorname{ar}(\Delta \mathrm{ABC}) \) 
Also \( (\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{ABC})-[\operatorname{ar}(\Delta \mathrm{BED})+\operatorname{ar}(\Delta \mathrm{CFE})+\operatorname{ar}(\Delta \mathrm{ADF})] \)
\( \Rightarrow \operatorname{ar}(\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{ABC})-\frac{3}{4} \operatorname{ar}(\Delta \mathrm{ABC})=\frac{1}{4} \operatorname{ar}(\Delta \mathrm{ABC}) \) 
\( \Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{DEF})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{1}{4} \) 

Ans : 
Let us assume two similar triangles as \( \triangle \mathrm{ABC} \sim \Delta \mathrm{PQR} \). Let AD and PS be the medians of these triangles.
\( \because \triangle A B C \sim \Delta P Q R \) 
\( \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}} \) ..(1)
\( \angle A=\angle P, \angle B=\angle Q, \angle C=\angle R \ldots(2) \) 
Since AD and PS are medians,
\( \therefore B D=D C=\frac{B C}{2} \) 
And, \( Q S=S R=\frac{Q R}{2} \) 
Equation (1) becomes
\( \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QS}}=\frac{\mathrm{AC}}{\mathrm{PR}} \) …(3)
In \( \triangle A B D \text { and } \Delta P Q S, \) 
\( \angle B=\angle Q[\text { Using equation }(2)] \) 
And, \( \frac{A B}{P Q}=\frac{B D}{Q S}[U s i n g \text { equation }(3)] \) 
\( \therefore \triangle \mathrm{ABD} \sim \Delta \mathrm{PQS} \) (SAS similarity criterion)
Therefore, it can be said that 
\( \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QS}}=\frac{\mathrm{AD}}{\mathrm{PS}} \) …(4)
\( \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)^{2} \) 
From equations (1) and (4), we may find that
\( \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}=\frac{A D}{P S} \) 
And hence, 
\( \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)^{2} \) 

Ans : 
Let ABCD be a square of side a.
Therefore, its diagonal = \( \sqrt{2} a \) 
Two desired equilateral triangles are formed as \( \triangle \mathrm{ABE} \text { and } \triangle \mathrm{DBF} \) 
Two desired equilateral triangles are formed as \( \triangle \mathrm{ABE} \text { and } \Delta \mathrm{DBF} \).
Side of an equilateral triangle, \( \triangle \mathrm{ABE} \), described on one of its sides = a
Side of an equilateral triangle, \( \triangle \mathrm{DBF} \) , described on one of its diagonals  \( =\sqrt{2} a \)
We know that equilateral triangles have all its angles as \( 60^{\circ} \) and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
\( \frac{\text { Area of } \Delta \mathrm{ABE}}{\text { Area of } \Delta \mathrm{DBF}}=\left(\frac{a}{\sqrt{2} a}\right)^{2}=\frac{1}{2} \) 

Ans : 
We know that equilateral triangles have all its angles as \( 60^{\circ} \) and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
Let side of \( \Delta \mathrm{ABC}=x \) 
Therefore, side of \( \Delta B D E=\frac{x}{2} \) 
\( \therefore \frac{\operatorname{area}(\Delta \mathrm{ABC})}{\operatorname{area}(\Delta \mathrm{BDE})}=\left(\frac{x}{\frac{x}{2}}\right)^{2}=\frac{4}{1} \) 
Hence, the correct answer is (C). 

Ans : If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.
It is given that the sides are in the ratio 4:9.
Therefore, ratio between areas of these triangles  \( =\left(\frac{4}{9}\right)^{2}=\frac{16}{81} \) 
Hence, the correct answer is (D). 

Ans : (i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides, we will obtain 49, 576, and 625.
49 + 576 = 625
Or, \( 7^{2}+24^{2}=25^{2} \) 
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
We know that the longest sides of a right triangle is the hypotenuse.
Therefore, the length of the hypotenuse of this triangle is 25 cm.

(ii) It is given that sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will obtain 9, 64, and 36.
However, 9 + 36 \( \neq 64 \) 
Or, \( 3^{2}+6^{2} \neq 8^{2} \) 
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Hence, it is not a right triangle
(iii) Given that sides are 50 cm, 80 cm. And 100 cm.
Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.
However, 2500 + 6400 \( \neq 10000 \) 
Or, \( 50^{2}+80^{2} \neq 100^{2} \) 
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.
(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will obtain 169, 144, and 24.
Clearly, 144 + 25 = 169.
Or, \( 12^{2}+5^{2}=13^{2} \) 
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
We know that the longest side of a right triangle is the hypotenuse.
Therefore, the length of the hypotenuse of this triangle is 13 cm. 

Ans : 
Let \( \angle \mathrm{MPR}=x \) 
In \( \Delta \mathrm{MPR} \) 
\( \begin{array}{l}{\angle M R P=180^{\circ}-90^{\circ}-x} \\ {\angle M R P=90^{\circ}-x}\end{array} \) 
\( \begin{aligned} \text { Similarly, in } \Delta \mathrm{MPQ} \\ \angle \mathrm{MPQ} &=90^{\circ}-\angle \mathrm{MPR} \\ &=90^{\circ}-x \end{aligned} \) 
\( \begin{array}{l}{\angle \mathrm{MQP}=180^{\circ}-90^{\circ}-\left(90^{\circ}-x\right)} \\ {\angle \mathrm{MOP}=x}\end{array} \) 
In \( \triangle \mathrm{QMP} \text { and } \Delta \mathrm{PMR} , \)
\( \begin{array}{l}{\angle \mathrm{MPQ}=\angle \mathrm{MRP}} \\ {\angle \mathrm{PMQ}=\angle \mathrm{RMP}} \\ {\angle \mathrm{MQP}=\angle \mathrm{MPR}}\end{array} \) 
\( \therefore \triangle \mathrm{QMP}-\mathrm{APMR} \quad(\text { By AAA similarity criterion }) \) 
\( \begin{array}{l}{\Rightarrow \frac{\mathrm{QM}}{\mathrm{PM}}=\frac{\mathrm{MP}}{\mathrm{MR}}} \\ {\Rightarrow \mathrm{PM}^{2}=\mathrm{QM} \times \mathrm{MR}}\end{array} \) 

Ans : (i) In \( \triangle \mathrm{ADB} \text { and } \triangle \mathrm{CAB} \), 
\( \angle D A B=\angle A C B \quad\left(\text { Each } 90^{\circ}\right) \) 
\( \angle \mathrm{ABD}=\angle \mathrm{CBA} \quad(\text { Common angle }) \) 
\( \therefore \triangle \mathrm{ADB}-\triangle \mathrm{CAB} \) (AA similarity criterion)
\( \Rightarrow \frac{A B}{C B}=\frac{B D}{A B} \) 
\( \Rightarrow A B^{2}=C B \times B D \) 

(ii) Let \( \angle \mathrm{CAB}=x \) 
In \( \Delta \mathrm{CBA} , \)
\( \begin{array}{l}{\angle C B A=180^{\circ}-90^{\circ}-x} \\ {\angle C B A=90^{\circ}-x}\end{array} \)
Similarly, In \( \triangle \mathrm{CAD} , \) 
\( \begin{aligned} \angle \mathrm{CAD} &=90^{\circ}-\angle \mathrm{CAB} \\ &=90^{\circ}-x \end{aligned} \) 
\( \begin{array}{l}{\angle C D A=180^{\circ}-90^{\circ}-\left(90^{\circ}-x\right)} \\ {\angle C D A=x}\end{array} \) 
In \( \triangle \mathrm{CBA} \text { and } \Delta \mathrm{CAD} \) 
\( \begin{aligned} \angle \mathrm{CBA} &=\angle \mathrm{CAD} \\ \angle \mathrm{CAB} &=\angle \mathrm{CDA} \end{aligned} \)
\( \angle A C B=\angle D C A \quad\left(\operatorname{Each} 90^{\circ}\right) \)
\( \therefore \Delta \mathrm{CBA} \sim \Delta \mathrm{CAD} \quad(\text { By AAA rule }) \) 
\( \Rightarrow \frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{BC}}{\mathrm{AC}} \) 
\( \Rightarrow \mathrm{AC}^{2}=\mathrm{DC} \times \mathrm{BC} \) 

(iii) In \( \triangle D C A \text { and } \Delta D A B, \) 
\( \begin{aligned} \angle \mathrm{DCA} &=\angle \mathrm{DAB}\left(\mathrm{Each} 90^{\circ}\right) \\ \angle \mathrm{CDA} &=\angle \mathrm{ADB}(\text { Common angle }) \end{aligned} \) 
\( \therefore \Delta \mathrm{DCA} \sim \Delta \mathrm{DAB} \quad \) (AA similarity criterion)
\( \Rightarrow \frac{\mathrm{DC}}{\mathrm{DA}}=\frac{\mathrm{DA}}{\mathrm{DB}} \) 
\( \Rightarrow A D^{2}=B D \times C D \) 

Ans : 
Given that \( \triangle \mathrm{ABC} \) is an isosceles triangle.
\( \therefore A C=C B \)
Applying Pythagoras theorem in \( \triangle \mathrm{ABC} \) (i.e., right-angled at point C), we obtain
\( \mathrm{AC}^{2}+\mathrm{CB}^{2}=\mathrm{AB}^{2} \) 
\( \Rightarrow \mathrm{AC}^{2}+\mathrm{AC}^{2}=\mathrm{AB}^{2} \) \( (A C=C B) \) 
\( \Rightarrow 2 \mathrm{AC}^{2}=\mathrm{AB}^{2} \) 

Ans : 
Given that,
\( \mathrm{AB}^{2}=2 \mathrm{AC}^{2} \) 
\( \Rightarrow \mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{AC}^{2} \) 
\( \Rightarrow \mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2} \quad(\mathrm{As \ } \mathrm{AC}=\mathrm{BC}) \)
The triangle is satisfying the pythagoras theorem.
Therefore, the given triangle is a right-angled triangle. 

Ans : 
Let AD be the altitude in the given equilateral triangle, \( \triangle \mathrm{ABC} \).
We know that altitude bisects the opposite side.
\( \therefore B D=D C=a \) 
In \( \triangle \mathrm{ADB} \) 
Applying pythagoras theorem, we obtain
\( \mathrm{AD}^{2}+\mathrm{DB}^{2}=\mathrm{AB}^{2} \)
\( \begin{array}{l}{\Rightarrow \mathrm{AD}^{2}+a^{2}=(2 a)^{2}} \\ {\Rightarrow \mathrm{AD}^{2}+a^{2}=4 a^{2}} \\ {\Rightarrow \mathrm{AD}^{2}=3 a^{2}} \\ {\Rightarrow \mathrm{AD}=a \sqrt{3}}\end{array} \) 
In an equilateral triangle, all the altitudes are equal in length.
Therefore, the length of each altitude will be \( \sqrt{3} a \). 

Ans : 
In \( \triangle \mathrm{AOB}, \Delta \mathrm{BOC}, \Delta \mathrm{COD}, \triangle \mathrm{AOD} \) 
Applying Pythagoras theorem, we obtain
\( \mathrm{AB}^{2}=\mathrm{AO}^{2}+\mathrm{OB}^{2} \) …(1)
\( \mathrm{BC}^{2}=\mathrm{BO}^{2}+\mathrm{OC}^{2} \) ….(2)
\( \mathrm{CD}^{2}=\mathrm{CO}^{2}+\mathrm{OD}^{2} \) …(3)
\( \mathrm{AD}^{2}=\mathrm{AO}^{2}+\mathrm{OD}^{2} \) …(4)
Adding all these equations, we obtain
\( \mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}+\mathrm{AD}^{2}=2\left(\mathrm{AO}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}+\mathrm{OD}^{2}\right) \) 
\( =2\left(\left(\frac{\mathrm{AC}}{2}\right)^{2}+\left(\frac{\mathrm{BD}}{2}\right)^{2}+\left(\frac{\mathrm{AC}}{2}\right)^{2}+\left(\frac{\mathrm{BD}}{2}\right)^{2}\right) \) 
( (Diagonals bisect each other) \) 
\( =2\left(\frac{(A C)^{2}}{2}+\frac{(B D)^{2}}{2}\right) \)
\( =(A C)^{2}+(B D)^{2} \) 

Ans : 
(i) Applying Pythagoras theorem in \( \triangle \mathrm{AOF} \), we obtain
\( \mathrm{OA}^{2}=\mathrm{OF}^{2}+\mathrm{AF}^{2} \) 
Similarly, in \( \triangle B O D \),
\( \mathrm{OB}^{2}=\mathrm{OD}^{2}+\mathrm{BD}^{2} \) 
Similarly, in \( \Delta \mathrm{COE} \),
\(  \mathrm{OC}^{2}=\mathrm{OE}^{2}+\mathrm{EC}^{2} \) 
Adding these equations,
\( \begin{array}{l}{\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}=\mathrm{OF}^{2}+\mathrm{AF}^{2}+\mathrm{OD}^{2}+\mathrm{BD}^{2}+\mathrm{OE}^{2}+\mathrm{EC}^{2}} \\ {\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{OD}^{2}-\mathrm{OE}^{2}-\mathrm{OF}^{2}=\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{EC}^{2}}\end{array} \) 
(ii) From the above result,
\( \begin{array}{l}{\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{EC}^{2}=\left(\mathrm{OA}^{2}-\mathrm{OE}^{2}\right)+\left(\mathrm{OC}^{2}-\mathrm{OD}^{2}\right)+\left(\mathrm{OB}^{2}-\mathrm{OF}^{2}\right)} \\ {\therefore \mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{EC}^{2}=\mathrm{AE}^{2}+\mathrm{CD}^{2}+\mathrm{BF}^{2}}\end{array} \) 

Ans : 
Let OA be the wall and AB be the ladder.
Therefore, by Pythagoras theorem,
\( \mathrm{AB}^{2}=\mathrm{OA}^{2}+\mathrm{BO}^{2} \) 
\( (10 \mathrm{m})^{2}=(8 \mathrm{m})^{2}+\mathrm{OB}^{2} \) 
\( 100 \mathrm{m}^{2}=64 \mathrm{m}^{2}+\mathrm{OB}^{2} \) 
\( \begin{array}{l}{\mathrm{OB}^{2}=36 \mathrm{m}^{2}} \\ {\mathrm{OB}=6 \mathrm{m}}\end{array} \) 
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m. 

Ans : 
Let OB be the pole and AB be the wire.
By Pythagoras theorem,
\( \mathrm{AB}^{2}=\mathrm{OB}^{2}+\mathrm{OA}^{2} \) 
\( (24 \mathrm{m})^{2}=(18 \mathrm{m})^{2}+\mathrm{OA}^{2} \) 
\( \mathrm{OA}^{2}=(576-324) \mathrm{m}^{2}=252 \mathrm{m}^{2} \) 
\( \mathrm{OA}=\sqrt{252} \mathrm{m}=\sqrt{6 \times 6 \times 7} \mathrm{m}=6 \sqrt{7} \mathrm{m} \) 
Therefore, the distance from the base is \( 6 \sqrt{7} \mathrm{m} \). 

Ans : 
Distance travelled by the plane flying towards north in \( 1 \frac{1}{2} \mathrm{hrs} \) 
\( =1,000 \times 1 \frac{1}{2}=1,500 \mathrm{km} \) 
Similarly, distance travelled vu the plane flying towards west in \( 1 \frac{1}{2} \mathrm{hrs} \) 
\( =1,200 \times 1 \frac{1}{2}=1,800 \mathrm{km} \)
Let these distances be represented by OA and OB respectively.
Applying Pythagoras as theorem,
Distance between these planes after \( 1 \frac{1}{2} h r s, A B=\sqrt{O A^{2}+O B^{2}} \) 
\( \begin{array}{l}{=\left(\sqrt{(1,500)^{2}+(1,800)^{2}}\right) \mathrm{km}=(\sqrt{2250000+3240000}) \mathrm{km}} \\ {=(\sqrt{5490000}) \mathrm{km}=(\sqrt{9 \times 610000}) \mathrm{km}=300 \sqrt{61} \mathrm{km}}\end{array} \)
Therefore, the distance between these planes will be \( 300 \sqrt{61} \mathrm{km} \) after \( 1 \frac{1}{2} \mathrm{hrs} \) 

Ans : 
Let CD and AB be the poles of height 11 m and 6 m.
Therefore, Cp = 11 - 6 = 5 m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for \( \triangle \mathrm{APC} \), we obtain
\( \begin{array}{l}{\mathrm{AP}^{2}+\mathrm{PC}^{2}=\mathrm{AC}^{2}} \\ {(12 \mathrm{m})^{2}+(5 \mathrm{m})^{2}=\mathrm{AC}^{2}} \\ {\mathrm{AC}^{2}=(144+25) \mathrm{m}^{2}=169 \mathrm{m}^{2}} \\ {\mathrm{AC}=13 \mathrm{m}}\end{array} \)
Therefore, the distance between their tops is 13 m. 

Ans : 
Applying Pythagoras theorem in \( \triangle \mathrm{ACE} \), we obtain
\( \mathrm{AC}^{2}+\mathrm{CE}^{2}=\mathrm{AE}^{2} \) …(1)
Applying Pythagoras theorem in \( \triangle B C D \) , we obtain
\( \mathrm{BC}^{2}+\mathrm{CD}^{2}=\mathrm{BD}^{2} \) ..(2)
Using equation (1) and equation (2), we obtain
\( \mathrm{AC}^{2}+\mathrm{CE}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}=\mathrm{AE}^{2}+\mathrm{BD}^{2} \) …(3)
Applying Pythagoras theorem in \( \Delta \mathrm{CDE} ,\) we obtain
\( \mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{CB}^{2} \) 
Putting the values in equation (3), we obtain
\( \mathrm{DE}^{2}+\mathrm{AB}^{2}=\mathrm{AE}^{2}+\mathrm{BD}^{2} \) 

Ans : Applying Pythagoras theorem for \( \triangle \mathrm{ACD} \), we obtain
\( \mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2} \)
\( \mathrm{AD}^{2}=\mathrm{AC}^{2}-\mathrm{DC}^{2} \) …(1)
Applying Pythagoras theorem in \( \triangle \mathrm{ABD} \), we obtain
\( \mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{DB}^{2} \) 
\( \mathrm{AD}^{2}=\mathrm{AB}^{2}-\mathrm{DB}^{2} \) …(2)
From equation (1) and equation (2), we obtain
\( \mathrm{AC}^{2}-\mathrm{DC}^{2}=\mathrm{AB}^{2}-\mathrm{DB}^{2} \quad \ldots \) (3)
It is given that 3DC = DB
\( \therefore \mathrm{DC}=\frac{\mathrm{BC}}{4} \text { and } \mathrm{DB}=\frac{3 \mathrm{BC}}{4} \) 
Putting these values in equation (3), we obtain,
\( \mathrm{AC}^{2}-\left(\frac{\mathrm{BC}}{4}\right)^{2}=\mathrm{AB}^{2}-\left(\frac{3 \mathrm{BC}}{4}\right)^{2} \) 
\( \mathrm{AC}^{2}-\frac{\mathrm{BC}^{2}}{16}=\mathrm{AB}^{2}-\frac{9 \mathrm{BC}^{2}}{16} \) 
\( \begin{array}{l}{16 \mathrm{AC}^{2}-\mathrm{BC}^{2}=16 \mathrm{AB}^{2}-9 \mathrm{BC}^{2}} \\ {16 \mathrm{AB}^{2}-16 \mathrm{AC}^{2}=8 \mathrm{BC}^{2}} \\ {2 \mathrm{AB}^{2}=2 \mathrm{AC}^{2}+\mathrm{BC}^{2}}\end{array} \) 

Ans : 
Let the side of the equilateral triangle be a, and AE be the altitude of \( \triangle \mathrm{ABC} \) 
\( \therefore B E=E C=\frac{B C}{2}=\frac{a}{2} \) 
And, \( A E=\frac{a \sqrt{3}}{2} \) 
Given that, BD = 1/3 BC
\( \therefore B D=\frac{a}{3} \) 
\( \mathrm{DE}=\mathrm{BE}-\mathrm{BD}= \) \( \frac{a}{2}-\frac{a}{3}=\frac{a}{6} \)
Applying Pythagoras theorem in \( \triangle \mathrm{ADE} \), we obtain
\( \mathrm{AD}^{2}=\mathrm{AE}^{2}+\mathrm{DE}^{2} \)
\( \mathrm{AD}^{2}=\left(\frac{a \sqrt{3}}{2}\right)^{2}+\left(\frac{a}{6}\right)^{2} \) 
\( =\left(\frac{3 a^{2}}{4}\right)+\left(\frac{a^{2}}{36}\right) \) 
\( =\frac{28 a^{2}}{36} \) 
\( =\frac{7}{9} \mathrm{AB}^{2} \) 
\( \Rightarrow 9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2} \) 

Ans : 
Let the side of the equilateral triangle be a, and AE be the altitude of \( \triangle \mathrm{ABC} \)
\( \therefore B E=E C=\frac{B C}{2}=\frac{a}{2} \) 
Applying Pythagoras theorem in \( \triangle \mathrm{ABE} \), we obtain
\( A B^{2}=A E^{2}+B E^{2} \)
\( a^{2}=\mathrm{AE}^{2}+\left(\frac{a}{2}\right)^{2} \) 
\( \mathrm{AE}^{2}=a^{2}-\frac{a^{2}}{4} \) 
\( \mathrm{AE}^{2}=\frac{3 a^{2}}{4} \) 
\( 4 \mathrm{AE}^{2}=3 \mathrm{a}^{2} \) 
\( \Rightarrow 4 \times(\text { Square of altitude })=3 \times \) (Square of one side) 

Ans : 
Given that, AB = \( =6 \sqrt{3} \mathrm{cm} \), AC = 12 cm, and BC = 6 cm
It can be observed that
\( \begin{array}{l}{A B^{2}=108} \\ {A C^{2}=144}\end{array} \) 
\( \begin{array}{l}{\text { And, } B C^{2}=36} \\ {A B^{2}+B C^{2}=A C^{2}}\end{array} \)
The given triangle, \( \triangle \mathrm{ABC} \), is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
\( \therefore \angle B=90^{\circ} \) 
Hence, the correct answer is (C) 

Ans : 
Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T,
Given that, PS is the angle bisector of \( \angle \mathrm{QPR} \).
\( \angle \mathrm{QPS}=\angle \mathrm{SPR} \ldots(1) \) 
By construction,
\( \angle S P R=\angle P R T(A S P S \| T R) \ldots(2) \)
\( \angle Q P S=\angle Q T R(A S P S| | T R) \ldots(3) \)
Using these equations, we obtain
\( \begin{array}{l}{\angle \mathrm{PRT}=\angle \mathrm{QTR}} \\ {\therefore \mathrm{PT}=\mathrm{PR}}\end{array} \) 
By construction,
PS || TR
By using basic proportionality theorem for \( \Delta \mathrm{QTR} \),
\( \frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{QP}}{\mathrm{PT}} \) 
\( \Rightarrow \frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}} \) (PT =TR) 

Ans : (i) Let us join DB.

We have, DN || CB, DM || AB, and \( \angle B=90^{\circ} \)
\( \therefore \mathrm{DMBN} \) is a rectangle.
\( \therefore \mathrm{DN}=\mathrm{MB} \text { and } \mathrm{DM}=\mathrm{NB} \)
The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.
\( \therefore \angle C D B=90^{\circ} \)
\( \Rightarrow \angle 2+\angle 3=90^{\circ} \ldots(1) \) 
In \( \triangle \mathrm{CDM} \),
\( \begin{array}{l}{\angle 1+\angle 2+\angle \mathrm{DMC}=180^{\circ}} \\ {\Rightarrow \angle 1+\angle 2=90^{\circ} \ldots(2)}\end{array} \)
In \( \triangle \mathrm{DMB} \),
\( \begin{array}{l}{\angle 3+\angle D M B+\angle 4=180^{\circ}} \\ {\Rightarrow \angle 3+\angle 4=90^{\circ} \ldots(3)}\end{array} \)
From equation (1) and (2), we obtain
\( \angle 2=\angle 4 \)
In \( \triangle \mathrm{D} \mathrm{CM} \text { and } \triangle \mathrm{BDM} \),
\( \begin{array}{l}{\angle 1=\angle 3 \text { (Proved above) }} \\ {\angle 2=\angle 4(\text { Proved above })}\end{array} \)
\( \therefore \triangle \mathrm{DCM} \sim \triangle \mathrm{BDM}(\text { AA similarity criterion) } \)
\( \Rightarrow \frac{\mathrm{BM}}{\mathrm{DM}}=\frac{\mathrm{DM}}{\mathrm{MC}} \) 
\( \Rightarrow \frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{DM}}{\mathrm{MC}} \quad(\mathrm{BM}=\mathrm{DN}) \) 
\( \Rightarrow \mathrm{DM}^{2}=\mathrm{D} \mathrm{N} \times \mathrm{MC} \)
(ii) In right triangle DBN,
\( \angle 5+\angle 7=90^{\circ} \ldots(4) \) 
\( In right triangle DAN, \)
\( \angle 6+\angle 8=90^{\circ} \ldots(5) \) 
D is the foot of the perpendicular drawn from B to AC.
\( \therefore \angle A D B=90^{\circ} \) 
\( \Rightarrow \angle 5+\angle 6=90^{\circ} \ldots(6) \)
From equation (5) and (6), we obtain
\( \angle 8=\angle 5 \) 
In \( \triangle \mathrm{DNA} \text { and } \triangle \mathrm{B} \mathrm{ND} \),
\( \begin{array}{l}{\angle 6=\angle 7 \text { (Proved above) }} \\ {\angle 8=\angle 5 \text { (Proved above) }}\end{array} \)
\( \therefore \triangle \mathrm{DNA} \sim \Delta \mathrm{BND} \) (AA similarity criterion)
\( \Rightarrow \frac{\mathrm{AN}}{\mathrm{DN}}=\frac{\mathrm{DN}}{\mathrm{NB}} \) 
\( \begin{array}{l}{\Rightarrow \mathrm{DN}^{2}=\mathrm{AN} \times \mathrm{NB}} \\ {\Rightarrow \mathrm{DN}^{2}=\mathrm{AN} \times \mathrm{DM}(\mathrm{AS} \mathrm{NB}=\mathrm{DM})}\end{array} \) 

Ans : Applying Pythagoras theorem in \( \triangle \mathrm{ADB} \), we obtain
\( A B^{2}=A D^{2}+D B^{2} \ldots(1) \)
Applying Pythagoras theorem in \( \triangle \mathrm{ACD} \), we obtain
\( \begin{array}{l}{A C^{2}=A D^{2}+D C^{2}} \\ {A C^{2}=A D^{2}+(D B+B C)^{2}} \\ {A C^{2}=A D^{2}+D B^{2}+B C^{2}+2 D B \times B C}\end{array} \)
\( A C^{2}=A B^{2}+B C^{2}+2 D B \times B C[\text { Using equation }(1)] \) 

Ans : Applying Pythagoras theorem in \( \triangle \mathrm{ADB} \), we obtain
\( \begin{array}{l}{A D^{2}+D B^{2}=A B^{2}} \\ {\Rightarrow A D^{2}=A B^{2}-D B^{2} \ldots(1)}\end{array} \) 
Applying Pythagoras theorem in \( \triangle \mathrm{ADC} \), we obtain
\( \mathrm{AD}^{2}+\mathrm{DC}^{2}=\mathrm{AC}^{2} \)
\( \begin{array}{l}{\mathrm{AB}^{2}-\mathrm{BD}^{2}+\mathrm{DC}^{2}=\mathrm{AC}^{2}[\text { Using equation }(1)]} \\ {\mathrm{AB}^{2}-\mathrm{BD}^{2}+(\mathrm{BC}-\mathrm{BD})^{2}=\mathrm{AC}^{2}} \\ {\mathrm{AC}^{2}=\mathrm{AB}^{2}+(\mathrm{BC}-\mathrm{BD})^{2}=\mathrm{AC}^{2}} \\ {\mathrm{AC}^{2}=\mathrm{AB}^{2}-\mathrm{BD}^{2}+\mathrm{BC}^{2}+\mathrm{BD}^{2}-2 \mathrm{BC} \times \mathrm{BD}} \\ {=\mathrm{AB}^{2}+\mathrm{BC}^{2}-2 \mathrm{BC} \times \mathrm{BD}}\end{array} \) 

Ans : (i) Applying Pythagoras theorem in \( \triangle \mathrm{AMD} \), we obtain
\( A M^{2}+M D^{2}=A D^{2} \ldots(1) \)
Applying Pythagoras theorem in \( \triangle \mathrm{AMC} \), we obtain
\( \begin{array}{l}{\mathrm{AM}^{2}+\mathrm{MC}^{2}=\mathrm{AC}^{2}} \\ {\mathrm{AM}^{2}+(\mathrm{MD}+\mathrm{DC})^{2}=\mathrm{AC}^{2}} \\ {\left(\mathrm{AM}^{2}+\mathrm{MD}^{2}\right)+\mathrm{DC}^{2}+2 \mathrm{MD} \cdot \mathrm{DC}=\mathrm{AC}^{2}}\end{array} \) 
\( A D^{2}+D C^{2}+2 M D . D C=A C^{2}[\text { Using equation }(1)] \)
Using the result, \( \quad \mathrm{DC}=\frac{\mathrm{BC}}{2}, \), we obtain
\( \mathrm{AD}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}+2 \mathrm{MD} \cdot\left(\frac{\mathrm{BC}}{2}\right)=\mathrm{AC}^{2} \)
\( \mathrm{AD}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}+\mathrm{MD} \times \mathrm{BC}=\mathrm{AC}^{2} \)

(ii) Applying Pythagoras theorem in \( \triangle A B M \), we obtain
\( A B^{2}=A M^{2}+M B^{2} \) 
\( \begin{array}{l}{=\left(A D^{2}-D M^{2}\right)+M B^{2}} \\ {=\left(A D^{2}-D M^{2}\right)+(B D-M D)^{2}} \\ {=A D^{2}-D M^{2}+B D^{2}+M D^{2}-2 B D \times M D} \\ {=A D^{2}+B D^{2}-2 B D \times M D}\end{array} \)
\( =\mathrm{AD}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}-2\left(\frac{\mathrm{BC}}{2}\right) \times \mathrm{MD} \) 
\( =\mathrm{AD}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}-\mathrm{BC} \times \mathrm{MD} \) 

(iii) Applying Pythagoras theorem in \( \triangle \mathrm{ABM} \), we obtain
\( A M^{2}+M B^{2}=A B^{2} \dots(1) \) 
Applying Pythagoras theorem in \( \triangle \mathrm{AMC} \), we obtain
\( \mathrm{AM}^{2}+\mathrm{MC}^{2}=\mathrm{AC}^{2} \ldots(2) \) 
Adding equations (1) and (2), we obtain
\( \begin{array}{l}{2 \mathrm{AM}^{2}+\mathrm{MB}^{2}+\mathrm{MC}^{2}=\mathrm{AB}^{2}+\mathrm{AC}^{2}} \\ {2 \mathrm{AM}^{2}+(\mathrm{BD}-\mathrm{DM})^{2}+(\mathrm{MD}+\mathrm{DC})^{2}=\mathrm{AB}^{2}+\mathrm{AC}^{2}} \\ {2 \mathrm{AM}^{2}+\mathrm{BD}^{2}+\mathrm{DM}^{2}-2 \mathrm{BD} \cdot \mathrm{DM}+\mathrm{MD}^{2}+\mathrm{DC}^{2}+\mathrm{AC}^{2}} \\ {2 \mathrm{AM}^{2}+2 \mathrm{MD}^{2}+\mathrm{BD}^{2}+\mathrm{DC}^{2}+2 \mathrm{MD}(-\mathrm{BD}+\mathrm{DC})=\mathrm{AB}^{2}+\mathrm{AC}^{2}}\end{array} \)
\( 2\left(\mathrm{AM}^{2}+\mathrm{MD}^{2}\right)+\left(\frac{\mathrm{BC}}{2}\right)^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}+2 \mathrm{MD}\left(-\frac{\mathrm{BC}}{2}+\frac{\mathrm{BC}}{2}\right)=\mathrm{AB}^{2}+\mathrm{AC}^{2} \) 
\( 2 \mathrm{AD}^{2}+\frac{\mathrm{BC}^{2}}{2}=\mathrm{AB}^{2}+\mathrm{AC}^{2} \) 

Ans : 
Let ABCD be a parallelogram.
Let us draw perpendicular DE on extended side AB, and AF on side DC.
Applying Pythagoras theorem in \( \triangle D E A \), we obtain
\( \mathrm{DE}^{2}+\mathrm{EA}^{2}=\mathrm{D} \mathrm{A}^{2} \ldots(i) \) 
Applying Pythagoras theorem in \( \triangle D E A \), we obtain
\( \mathrm{DE}^{2}+\mathrm{EA}^{2}=\mathrm{DA}^{2} \ldots \) (i)
Applying Pythagoras theorem in \( \triangle \mathrm{DEB} \), we obtain
\( \begin{array}{l}{\mathrm{DE}^{2}+\mathrm{EB}^{2}=\mathrm{DB}^{2}} \\ {\mathrm{DE}^{2}+(\mathrm{EA}+\mathrm{AB})^{2}=\mathrm{DB}^{2}} \\ {\left(\mathrm{DE}^{2}+\mathrm{EA}^{2}\right)+\mathrm{AB}^{2}+2 \mathrm{EA} \times \mathrm{AB}=\mathrm{DB}^{2}} \\ {\mathrm{DA}^{2}+\mathrm{AB}^{2}+2 \mathrm{E} \mathrm{A} \times \mathrm{AB}=\mathrm{DB}^{2} \ldots \text { (ii) }}\end{array} \)
Applying Pythagoras theorem in \( \triangle \mathrm{ADF} \), we obtain
\( \mathrm{AD}^{2}=\mathrm{AF}^{2}+\mathrm{FD}^{2} \)
Applying Pythagoras theorem in \( \triangle \mathrm{AFC} \), we obtain
\( A C^{2}=A F^{2}+F C^{2} \)
\( \begin{array}{l}{=\mathrm{AF}^{2}+(\mathrm{DC}-\mathrm{FD})^{2}} \\ {=\mathrm{AF}^{2}+\mathrm{DC}^{2}+\mathrm{FD}^{2}-2 \mathrm{DC} \times \mathrm{FD}} \\ {=\left(\mathrm{AF}^{2}+\mathrm{FD}^{2}\right)+\mathrm{DC}^{2}-2 \mathrm{DC} \times \mathrm{FD}} \\ {\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}-2 \mathrm{DC} \times \mathrm{FD} \ldots(i i)}\end{array} \) 
Since ABCD is a parallelogram,
\( \begin{array}{l}{A B=C D \ldots(i v)} \\ {\text { And, } B C=A D \ldots(v)}\end{array} \) 
In \( \triangle \mathrm{DE} \mathrm{A} \text { and } \triangle \mathrm{ADF} \)
\( \begin{aligned} \angle D E A &=\angle A F D\left(B o t h 90^{\circ}\right) \\ \angle E A D &=\angle A D F(E A \| D F) \end{aligned} \) 
\( A D=A D(\text { Common }) \) 
\( \therefore \Delta E A D \cong \triangle F D A \) (AAS congruence criterion)
\( \Rightarrow E A=D F \ldots(v i) \) 
Adding equations (i) and (iii), we obtain
\( \begin{array}{l}{\mathrm{DA}^{2}+\mathrm{AB}^{2}+2 \mathrm{EA} \times \mathrm{AB}+\mathrm{AD}^{2}+\mathrm{DC}^{2}-2 \mathrm{DC} \times \mathrm{FD}=\mathrm{DB}^{2}+\mathrm{AC}^{2}} \\ {\mathrm{DA}^{2}+\mathrm{AB}^{2}+\mathrm{AD}^{2}+\mathrm{DC}^{2}+2 \mathrm{EA} \times \mathrm{AB}-2 \mathrm{DC} \times \mathrm{FD}=\mathrm{DB}^{2}+\mathrm{AC}^{2}} \\ {\mathrm{BC}^{2}+\mathrm{AB}^{2}+\mathrm{AD}^{2}+\mathrm{DC}^{2}+2 \mathrm{E} \mathrm{A} \times \mathrm{AB}-2 \mathrm{AB} \times \mathrm{EA}=\mathrm{DB}^{2}+\mathrm{AC}^{2}}\end{array} \)
[Using equations (iv) and (vi)]
\( A B^{2}+B C^{2}+C D^{2}+D A^{2}=A C^{2}+B D^{2} \) 

Ans : 
(i) In \( \triangle A P C \text { and } \Delta D P B \),
\( \angle \mathrm{APC}=\angle \mathrm{DPB} \) (Vertically opposite angles)
\( \angle \mathrm{CAP}=\angle \mathrm{BDP} \) ( Angles in the same segment for chord CB)
\( \triangle \mathrm{APC} \sim \Delta \mathrm{DPB} \) (By AA similarity criterion)
(ii) We have already proved that
\( \triangle \mathrm{APC} \sim \Delta \mathrm{DPB} \)
We know that the corresponding sides of similar triangles are proportional.
\( \therefore \frac{\mathrm{AP}}{\mathrm{DP}}=\frac{\mathrm{PC}}{\mathrm{PB}}=\frac{\mathrm{CA}}{\mathrm{BD}} \) 
\( \Rightarrow \frac{A P}{D P}=\frac{P C}{P B} \) 
\( \therefore \mathrm{AP} . \mathrm{PB}=\mathrm{PC} . \mathrm{DP} \) 

Ans : (i) In \( \triangle \mathrm{PAC} \text { and } \Delta \mathrm{PDB} \),
\( \angle P=\angle P(\text { Common }) \) 
\( \begin{array}{l}{\angle \mathrm{PAC}=\angle \mathrm{PDB} \text { (Exterior angle of a cyclic quadrilateral is } \angle \mathrm{PCA}=\angle \mathrm{PBD} \text { equal to the }} \\ {\text { opposite interior angle) }}\end{array} \)
\( \therefore \triangle \mathrm{PAC} \sim \Delta \mathrm{PDB} \) 
(ii) We know that the corresponding sides of similar triangles are proportional.
\( \therefore \frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{AC}}{\mathrm{DB}}=\frac{\mathrm{PC}}{\mathrm{PB}} \) 
\( \Rightarrow \frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PB}} \) 
\( \therefore \text { PA.PB }=\mathrm{PC.PD} \) 

Ans : Let us extend BA to P such that AP =AC, Join PC.

It is given that,
\( \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}} \) 
\( \Rightarrow \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AP}}{\mathrm{AC}} \) 
By using the converse of basic proportionality theorem, we obtain AD || PC
\( \mathrm{AD}\|\mathrm{PC} \) 
\( \Rightarrow \angle B A D=\angle A P C(\text { Corresponding angles }) \ldots \) (1)
And \( \angle D A C=\angle A C P \) (Alternate interior angles)...(2)

By construction, we have
\( A P=A C \) 
\( \Rightarrow \angle A P C=\angle A C P \ldots(3) \) 
On comparing equations (1), (2), and (3), we obtain
\( \angle B A D=\angle A P C \)
\( \Rightarrow \mathrm{AD} \) is the bisector of the angle BAC. 

Ans : 
Let AB be the height of the tip of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod.
Then, AC is the length of the string.
AC can be found by applying Pythagoras theorem in \( \triangle \mathrm{ABC} \)
\( A C^{2}=A B^{2}+B C^{2} \) 
\( A B^{2}=(1.8 m)^{2}+(2.4 m)^{2} \)
\( \begin{array}{l}{A B^{2}=(3.24+5.76) m^{2}} \\ {A B^{2}=9.00 m^{2}}\end{array} \)
\( \Rightarrow A B=\sqrt{9} \mathrm{m}=3 \mathrm{m} \)
Thus, the length of the string out is 3 m.
She pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m.