NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals – Here are all the NCERT solutions for Class 9 Maths Chapter 8. This solution contains questions, answers, images, explanations of the complete Chapter 8 titled The Fun They Had of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 8 Quadrilaterals. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals in one place.

## NCERT Solutions Class 9 Maths Chapter 8 QUADRILATERALS

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 8 QUADRILATERALS , Maths, Class 9.

 Class 9 Subject Maths Book Mathematics Chapter Number 8 Chapter Name QUADRILATERALS

### NCERT Solutions Class 9 Maths chapter 8 QUADRILATERALS

Class 9, Maths chapter 8, QUADRILATERALS solutions are given below in PDF format. You can view them online or download PDF file for future use.

Q.1: The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Ans : Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x, 9x, and 13x respectively. As the sum Of all interior angles Of a quadrilateral is $$360^{\circ}$$, Therefore, 3x + 5x + 9x + 13x = $$360^{\circ}$$ 30x = $$360^{\circ}$$ x = $$12^{\circ}$$ Hence, the angles are $$\begin{array}{l}{3 x=3 \times 12=360} \\ {5 x=5 \times 12=60^{\circ}} \\ {9 x=9 \times 12=108^{\circ}} \\ {13 x=13 \times 12=156^{\circ}}\end{array}$$

Q.2: If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Ans : Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is $$90^{\circ}$$. In $$\triangle \mathrm{A} \mathrm{BC}$$ and $$\triangle \mathrm{D} \mathrm{CB}$$, AB = DC (Opposite sides Of a parallelogram are equal) BC = BC (Common) AC = DB (Given) $$\therefore \triangle A B C \cong \triangle D C B$$ (By SSS Congruence rule) $$\Rightarrow \angle A B C=\angle D C B$$ It is known that the sum of the measures of angles on the same side of transversal is $$180^{\circ}$$. $$\begin{array}{l}{\angle A B C+\angle D C B=180^{\circ}(A B \| C D)} \\ {\Rightarrow \angle A B C+\angle A B C=180^{\circ}} \\ {\Rightarrow 2 \angle A B C=180^{\circ}} \\ {\Rightarrow \angle A B C=90^{\circ}}\end{array}$$ Since ABCD is a parallelogram and one of its interior angles is $$90^{\circ}$$ , ABCD is a rectangle.

Q.3: Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Ans : Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each Other at right angle i.e., OA = OC, OB = OD, and $$\angle A O B=\angle B O C=\angle C O D=\angle A O D=90^{\circ}$$. To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal. In $$\triangle \mathrm{AOD} \text { and } \Delta \mathrm{COD}$$ , OA = OC (Diagonals bisect each other ) $$\angle A O D=\angle C O D$$ (Given) OD = OD (Common) $$\begin{array}{l}{\therefore \triangle A O D=\Delta C O D(B y \text { SAS congruence rule) }} \\ {\therefore A D=C D(1)}\end{array}$$ Similarly, it can be proved that AD = AB and CD = BC (2) From equations (1) and (2) , AB = BC = CD = AD Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a rhombus.

Q.4: Show that the diagonals of a square are equal and bisect each other at right angles.

Ans : Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O. To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove AC = BD, OA = OC, OB = OD, and $$\angle A O B=90^{\circ}$$ . In $$\triangle A B C \text { and } \Delta D C B$$, AB = DC (Sides Of a square are equal to each Other) $$\angle A B C=\angle D C B$$ (All interior angles are of $$90^{\circ}$$ ) BC = CB (Common side) $$\begin{array}{l}{\therefore \triangle \mathrm{ABC}=\Delta \mathrm{DCB}(\mathrm{By} \text { SAS congruency })} \\ {\therefore \mathrm{AC}=\mathrm{DB}(\mathrm{By} \mathrm{CPCT})}\end{array}$$ Hence, the diagonals of a square are equal in length. In $$\triangle \mathrm{AOB} \text { and } \Delta \mathrm{COD}$$ , $$\angle A O B=\angle C O D$$ (Vertically Opposite Angle ) $$\square A B O=\square C D O$$ (Alternate interior angles) AB = CD (Sides of a square are always equal ) $$\begin{array}{l}{\square \triangle A O B \square \Delta C O D(B y \text { AAS congruence rule) }} \\ {\square A O=C O \text { and } O B=O D(B y C P C T)}\end{array}$$ Hence, the diagonals of a square bisect each other. In $$\triangle A O B \text { and } \Delta C O B$$ , As we had proved that diagonals bisect each other , therefore , AO = CO AB = CB (Sides of a square are equal) BO = BO (Common) $$\begin{array}{l}{\square \triangle A O B \square \Delta C O B(B y \text { SSS congruence rule) }}\end{array}$$ $$\square \square A O B=\square C O B(B y C P C T)$$ However, $$\square A O B+\square C O B=180^{\circ}$$ (Linear pair ) $$\begin{array}{l}{2 \square A O B=180^{\circ}} \\ {\square A O B=90^{\circ}}\end{array}$$ Hence, the diagonals of a square bisect each other at right angles.

Q.5: Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Ans : Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC = BD, OA = OC, 0B = OD, and $$\angle A O B=\angle B O C=\angle C O D$$$$=\angle A O D=90^{\circ}$$. To prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is $$90^{\circ}$$ . In $$\triangle \mathrm{AOB} \text { and } \Delta \mathrm{COD}$$, AO = CO (Diagonals bisect each other) OB = OD (Diagonals bisect each other) $$\angle A O B=\angle C O D(\text { Vertically opposite angles })$$ $$\square \triangle \mathrm{AOB} = \square \Delta \mathrm{COD}(\text { SAS congruence rule) }$$ $$\begin{array}{l}{\square A B=C D(B y C P C T) \ldots(1)} \\ {\text { And, } \square O A B=\square O C D(B y C P C T)}\end{array}$$ However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel. $$\square A B\|C D \ldots(2)$$ From equations (1) and (2), we obtain ABCD is a parallelogram. In $$\triangle \mathrm{AOD} \text { and } \triangle \mathrm{COD}$$ , AO = CO (diagonals bisect each other) $$\square A O D=\square C O D\left(\text { Given that each is } 90^{\circ}\right)$$ OD = OD (Common) $$\begin{array}{l}{\square \triangle A O D \square \Delta C O D(S A S \text { congruence rule) }} \\ {\square A D=D C \ldots(3)}\end{array}$$ However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD) $$\square A B=B C=C D=D A$$ Therefore, all the sides of quadrilateral ABCD are equal to each other. In $$\triangle A D C \text { and } \Delta B C D,$$ , AD = BC (Already proved) AC = BD (Given) DC =CD (Common) $$\begin{array}{l}{\square \triangle \mathrm{ADC} \square \Delta \mathrm{BCD}(\mathrm{SSS} \text { Congruence rule) }} \\ {\square \square \mathrm{ADC}=\square \mathrm{BCD}(\mathrm{By} \mathrm{CPCT})}\end{array}$$ However, $$\square A D C+\square B C D=180^{\circ}(\text { Co-interior angles })$$ $$\begin{array}{l}{\square \square A D C+\square A D C=180^{\circ}} \\ {\square 2 \square A D C=180^{\circ}} \\ {\square \square A D C=90^{\circ}}\end{array}$$ One of the interior angles of quadrilateral ABCD is a right angle. Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is $$90^{\circ}$$ .Therefore, ABCD is a square.

Q.6: Diagonal AC of a parallelogram ABCD bisects ∠ A
Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus.

Ans : (i) ABCD is a parallelogram. $$\square \square D A C=\square B C A$$ (Alternate interior angles ) …..(1) And, $$\square B A C=\square D C A$$ (Alternate interior angles ) …..(2) However, it is given that AC bisects $$\square \mathrm{A}$$ . $$\square \square D A C=\square B A C \ldots(3)$$ From equations (1), (2), and (3), we obtain $$\begin{array}{l}{\square \mathrm{DAC}=\square B C A=\square B A C=\square D C A \ldots(4)} \\ {\square \square D C A=\square B C A}\end{array}$$ Hence, AC bisects $$\square \mathrm{C}$$ . (ii) From equation (4) , we obtain $$\begin{array}{l}{\square D A C=\square D C A} \\ {\square D A=D C(\text { Side opposite to equal angles are equal) }}\end{array}$$ However, DA = BC and AB = CD (Opposite sides of a parallelogram) $$\square A B=B C=C D=D A$$ Hence, ABCD is a rhombus.

Q.7: ABCD is a rhombus. Show that diagonal AC bisects ∠ A as well as ∠ C and diagonal BD bisects ∠ B as well as ∠ D.

Ans : Let us join AC. In $$\triangle \mathrm{ABC}$$ , BC = AB (Sides of a rhombus are equal to each other ) ∠1= ∠2 (Angles opposite to equal sides of a triangle are equal) However, ∠1=∠3 (Alternate interior angles for parallel lines AB and CD) ∠2 = ∠3 Therefore, AC bisects ∠C. Also, ∠2 = ∠4 (Alternate interior angles for parallel lines BC and DA ∠1 = ∠4 Therefore, AC bisects ∠A Similarly, it can be proved that BD bisects ∠B and ∠D as well.

Q.8: ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠ B as well as ∠ D.

Ans : (i) It is given that ABCD is a rectangle. $$\begin{array}{l}{\angle A=\angle C} \\ {\Rightarrow \frac{1}{2} \angle A=\frac{1}{2} \angle C} \\ {\Rightarrow \angle D A C=\angle D C A}\end{array}$$ $$(\mathrm{AC} \text { bisects } \angle \mathrm{A} \text { and } \angle \mathrm{C})$$ CD = DA (Sides opposite to equal angles are also equal) However, DA = BC and AB = CD (Opposite sides of a rectangle are equal) $$\square A B=B C=C D=D A$$ ABCD is a rectangle and all of its sides are equal. Hence, ABCD is a square. (ii) Let us join BD. In $$\Delta B C D$$ , BC = CD (Sides of a square are equal to each other.) $$\angle C D B=\angle C B D$$ (Angles opposite to equal sides are equal) However, $$\angle \mathrm{CDB}=\angle \mathrm{ABD} \text { (Alternate interior angles for } \mathrm{AB} \| \mathrm{CD} )$$ $$\begin{array}{l}{\triangle C B D=\triangle A B D} \\ {B D \text { bisects } \angle B}\end{array}$$ Also, $$\triangle \mathrm{CBD}=\triangle \mathrm{ADB}$$ (Alternate interior angles for BC II AD ) $$\begin{array}{l}{\angle C D B=\angle A B D} \\ { B D \text { bisects } \angle D}\end{array}$$

Q.9: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ

Show that:
(i) ∆ APD ≅ ∆ CQB .
(ii) AP = CQ
(iii) ∆ AQB ≅∆ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.

Ans : (i) In $$\triangle A P D \text { and } \Delta C Q B,$$ , $$\square A D P=\square C B Q$$ (Alternate interior angles for BC II AD ) AD = CB (Opposite sides of a parallelogram ABCD) DP = BQ (Given ) $$\square \triangle A P D \square \Delta C Q B$$ (Using SAS Congruence rule) (ii) As we have observed that $$\triangle \mathrm{APD} \square \Delta \mathrm{CQB}$$, $$\square A P=C Q(C P C T)$$ (iii) In $$\triangle \mathrm{AQB} \text { and } \Delta \mathrm{CPD}$$ $$\square A B Q=\square C D P(\text { Alternate interior angles for } A B \| C D)$$ AB = CD (Opposite sides of a parallelogram ABCD ) BQ = DP (Given) $$\square \triangle A Q B \square \Delta C P D(U \text {sing SAS congruence rule) }$$ (iv) As we had observed that $$\triangle \mathrm{AQB} \square \Delta \mathrm{CPD}$$, $$\square A Q=C P(C P C T)$$ (v) From the result obtained in (ii) and (iv) , AQ = CP and AP = CQ Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram.

Q.10: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD
.
Show that
(i) ∆ APB ≅ ∆ CQD
(ii) AP = CQ

Ans : (i) ∆ APB and ∆ CQD $$\square A P B=\square C Q D\left(E a c h 90^{\circ}\right)$$ AB = CD (Opposite sides of a parallelogram ABCD ) $$\square A B P=\square C D Q$$ (Alternate interior angles for AB II CD ) $$\square \triangle \mathrm{APB} \square \Delta \mathrm{CQD}(\mathrm{By} \text { AAS congruence })$$ (ii) By using the above result $$\triangle A P B \square \Delta C Q D$$ , we obtain AP = CQ (By CPCT)

Q.11: In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively
.
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ ABC ≅ ∆ DEF.

Ans : It is given that AB = DE and AB II DE. If two opposite sides of a quadrilateral are equal and parallel to each other , then it will be a parallelogram. Therefore, quadrilateral ABED is a parallelogram. (ii) Again, BC = EF and BC II EF Therefore, quadrilateral BCEF is a parallelogram. (iii) As we had observed that ABED and BEFC are parallelograms, therefore AD = BE and AD II BE (Opposite sides of a parallelogram are equal and parallel) And, BE = CF and BE II CF (Opposite sides of a parallelogram are equal and parallel) $$\square A D=C F \text { and } A D\|C F$$ (iv) As we had observed that one pair of opposite sides (AD and CF ) of a quadrilateral ACFD are equal and parallel to each other, therefore , it is a parallelogram. (v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other. $$\square A C\|D F$$ and AC = DF (vi) $$\triangle \mathrm{ABC} \text { and } \Delta \mathrm{DEF}$$ , AB = DE ( Given ) BC = EF (Given) AC = DF (ACFD is parallelogram) $$\square \triangle A B C \square \Delta D E F$$ ( By SSS Congruence Rule)

Q.12: ABCD is a trapezium in which AB || CD and AD = BC
.
Show that
(i) ∠ A = ∠ B
(ii) ∠ C = ∠ D
(iii) ∆ ABC ≅ ∆ BAD
(iv) diagonal AC = diagonal BD
[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Ans : Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. (i) AD = CE (Opposite sides of a parallelogram AECD) However, AD = BC (Given) Therefore, BC = CE $$\square C E B=\square C B E$$ ( Angle opposite to equal sides are also equal) Consider parallel lines AD and CE. AE is the transversal line for them. $$\square A+\square C E B=180^{\circ}$$ (Angles on the same side of transversal ) $$\square A+\square C B E=180^{\circ}(\text { Using the relation } \square C E B=\square C B E) \ldots(1)$$ However, $$\square B+\square C B E=180^{\circ}(\text { Linear pair angles }) \dots .$$ (2) From equations (1) and (2) , we obtain $$\square A=\square B$$ (ii) AB II CD $$\square A+\square D=180^{\circ}$$ (Angles on the same side are transversal) Also, $$\square C+\square B=180^{\circ}$$ (Angles on the same side are transversal) $$\square \square A+\square D=\square C+\square B$$ However, $$\square A=\square B$$ [ Using the result obtained in (i) ] $$\square \square C=\square D$$ (iii) In $$\triangle A B C \text { and } \Delta B A D,$$ , AB = BA (Common side ) BC = AD (Given) $$\begin{array}{l}{\square B=\square A(\text { Proved before })} \\ {\square \triangle A B C \square \Delta B A D \text { (SAS congruence rule) }}\end{array}$$ (iv) We had observed that, $$\begin{array}{l}{\triangle \mathrm{ABC} \square \Delta \mathrm{BAD}} \\ {\square \mathrm{AC}=\mathrm{BD}(\mathrm{By} \mathrm{CPCT})}\end{array}$$

Q.1: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA
.
AC is a diagonal. Show that :
(i) SR || AC and SR = $$\frac{1}{2}$$ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Ans : (i) In $$\triangle \mathrm{ADC}$$, S and R are the mid-points of sides AD and CD respectively. In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the 3rd side and is half of it. $$\square \text { SR }\left\|A C \text { and } S R=\frac{1}{2} A C \ldots(1)\right.$$ (ii) In $$\triangle \mathrm{ABC}$$ . P and Q are the mid-points of sides AB and BC respectively. Therefore, by using mid-point theorem, $$\mathrm{PQ}\left\|\mathrm{AC} \text { and } \mathrm{PQ}=\frac{1}{2} \mathrm{AC} \ldots(2)\right.$$ Using equations (1) and (2) , we obtain PQ II SR and PQ = SR ….. (3) $$\square \mathrm{PQ}=\mathrm{SR}$$ (iii) From equation (3), we obtained PQ II SR and PQ = SR Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. Hence, PQRS is a parallelogram.

Q.2: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Ans : and S are the mid-points of sides CD and AD respectively. $$\square \mathrm{RS}\left\|\mathrm{AC} \text { and } \mathrm{RS}=\frac{1}{2} \mathrm{AC}(\text { Using mid-point theorem }) \ldots \text { (2) }\right.$$ From equations (1) and (2) , we obtain PQ II RS and PQ = RS Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram. Let the diagonals of rhombus ABCD intersect each other at point O. In quadrilateral OMPQ, mq II ON ( $$(\because \mathrm{PQ} \| \mathrm{AC})$$ qn II OM ( $$(\because \mathrm{QR} \| \mathrm{BD})$$ Therefore, OMQN is a parallelogram. $$\begin{array}{l}{\square \square M Q N=\square N O M} \\ {\square \square P Q R=\square N O M}\end{array}$$ However, $$\square \mathrm{NOM}=90^{\circ}$$ (Diagonals of a rhombus are perpendicular to each other). $$\square \square P Q R=90^{\circ}$$ Clearly, PQRS is a parallelogram having one of its interior angles as $$90^{\circ}$$. Hence, PQRS is a rectangle.

Q.3: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Ans : Let us join AC and BD. In $$\triangle \mathrm{ABC}$$ , P and Q are the mid-points of AB and BC respectively. $$\square \mathrm{PQ}\left\|\mathrm{AC} \text { and } \mathrm{PQ}=\frac{1}{2} \mathrm{AC}(\text { Mid-point theorem) } \ldots \text { (1) }\right.$$ Similarly in $$\triangle \mathrm{ADC}$$, $$\mathrm{SR}\left\|\mathrm{AC} \text { and } \mathrm{SR}=\frac{1}{2} \mathrm{AC}(\text { Mid-point theorem }) \ldots \text { (2) }\right.$$ Clearly, PQ II SR and PQ = SR Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram. $$\square \text { PS }\|\text { QR and } \mathrm{PS}=\text { QR (Opposite sides of parallelogram)... (3) }$$ In $$\triangle B C D$$ , Q and R are the mid-points of side BC and CD respectively. $$\square \mathrm{QR}\left\|\mathrm{BD} \text { and } \mathrm{QR}=\frac{1}{2} \mathrm{BD}\right.$$ (Mid-point theorem) …… (4) However,the diagonals of a rectangle are equal. $$\square A C=B D \ldots(5)$$ By using equation (1), (2), (3), (4), and (5), we obtain PQ = QR = SR = PS Therefore, PQRS is a rhombus.

Q.4: ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F.
.
Show that F is the midpoint of BC.

Ans : Let EF intersect DB at G. By converse of midpoint theorem, we know that a line drawn through the midpoint of any side of a triangle and parallel to another side, bisects the third side. In $$\triangle \mathrm{ABD}$$, EF II AB and E is the midpoint of AD. Therefore, G will be the midpoint of DB. As EF II AB and AB II CD , $$\square \mathrm{EF}\|\mathrm{CD}$$ (Two lines parallel to the same line are parallel to each other.) In $$\triangle B C D$$ , GF II CD and G is the midpoint of line BD. Therefore, by using converse of mid-point theorem, F is the midpoint of BC.

Q.5: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively
.
Show that the line segments AF and EC trisect the diagonal BD.

Ans : ABCD is a parallelogram. $$\square A B\|C D$$ And hence, AE II FC Again, AB = CD (Opposite sides of a parallelogram ABCD) $$\frac{1}{2} A B=\frac{1}{2} C D$$ AE = FC ( E and F are mid-points of side AB and CD) In quadrilateral AECF, one pair of opposite sides (AE and CF ) is parallel and equal to each other. Therefore, AECF is a parallelogram. $$\square \text { AF }\|\mathrm{EC}$$ (Opposite sides of a parallelogram) In $$\Delta \mathrm{DQC}$$, F is the midpoint of side DC and FP II CQ (as AF II EC). Therefore, by using converse of midpoint theorem , it can be said that P is the mid-point of DQ. $$\square \mathrm{DP}=\mathrm{PQ} \ldots(1)$$ Similarly, in $$\triangle \mathrm{APB}$$, E is the midpoint of side AB and EQ II AP (as AF II EC ). Therefore, by using converse of midpoint theorem, it can be said that Q is the mid-point of PB. $$\square P Q=Q B \ldots(2)$$ From equations (1) and (2), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD.

Q.6: Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Ans : Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. JOin PQ, QR, RS, SP, and BD. In $$\triangle A B D$$, S and P are the mid-points Of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that $$\mathrm{SP}\left\|\mathrm{BD} \text { and } \mathrm{SP}=\frac{1}{2} \mathrm{BD} \ldots(1)\right.$$ Similarly in $$\triangle B C D$$, $$\mathrm{QR}\left\|\mathrm{BD} \text { and } \mathrm{QR}=\frac{1}{2} \mathrm{BD} \ldots(1)\right.$$ $$\begin{array}{l}{\text { From equations }(1) \text { and }(2), \text { we obtain }} \\ {\text { SP }\|\text { QR and } S P=Q R}\end{array}$$ In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other. Therefore, SPQR is a parallelogram. We know that diagonals Of a parallelogram bisect each other. Hence, PR and QS bisect each other.

Q.7: ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the midpoint of AC
(ii) MD ⊥AC
(iii) CM = MA = $$\frac{1}{2}$$ AB

Ans : (i) In $$\triangle \mathrm{ABC}$$, It is given that M is the midpoint of AB and MD II BC. Therefore, D is the midpoint of AC. (Converse of mid-point theorem) (ii) As DM II CB and AC is a transversal line for them, therefore, $$\begin{array}{l}{\square M D C+\square D C B=180^{\circ}(C o-\text { interior angles })} \\ {\square M D C+90^{\circ}=180^{\circ}} \\ {\square M D C=90^{\circ}} \\ {\square M D \square A C}\end{array}$$ (iii) Join MC. In $$\triangle \mathrm{AMD} \text { and } \Delta \mathrm{CMD}$$, AD = CD (D is the midpoint Of side AC) $$\square A D M=\square C D M\left(E a c h 90^{\circ}\right)$$ DM = DM (Common) $$\square \triangle A M D \square \Delta C M D$$ (By SAS Congruence Rule) Therefore, AM = CM (By CPCT) However, AM = $$\frac{1}{2}_{A B}$$ (M is the midpoint of AB) Therefore, it can be said that $$\mathrm{CM}=\mathrm{AM}=\frac{1}{2} \mathrm{AB}$$

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