NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Ans : In ∆ ABC and ∆ ABD,
AC = AD (Given)
∠CAB = ∠ DAB (AB bisects ∠A)
AB = AB (Common)
Therefore,  ∆ ABC ≅ ∆ ABD( By SAS congruence rule)
BC = BD (By CPCT)
Therefore, BC and BD are Of equal lengths. 

Ans : In ∠ABD and ∠BAC,
AD = BC (Given)
∠DAB = ∠CBA(Given)
AB = BA (Common)
\(\begin{array}{l}{\therefore \triangle \mathrm{ABD}=\Delta \mathrm{BAC}(\mathrm{By} \text { SAS congruence rule) }} \\ {\therefore \mathrm{BD}=\mathrm{AC}(\mathrm{By} \mathrm{CPCT})}\end{array}\)
And, ∠ABD = ∠BAC (By CPCT) 

Ans : In ∆ BOC and ∆ AOD,
\(\begin{array}{l}{\angle \mathrm{BOC}=\angle \mathrm{AOD}(\text { Vertically opposite angles })} \\ {\angle \mathrm{CBO}=\angle \mathrm{DAO}\left(\text { Each } 90^{\circ}\right)}\end{array}\)
\(\begin{array}{l}{\mathrm{BC}=\mathrm{AD}(\text { Given })} \\ {\therefore \Delta \mathrm{BOC}=\triangle \mathrm{AOD} \text { (AAS congruence rule) }} \\ {\therefore \mathrm{BO}=\mathrm{AO}(\mathrm{By} \mathrm{CPCT})} \\ {\Rightarrow \mathrm{CD} \text { bisects } \mathrm{AB}}\end{array}\) 

Ans : In ∆ ABC and  ∆ CDA, \(\begin{array}{l}{\angle B A C=\angle D C A(\text { Alternate interior angles, as } p \| q)} \\ {A C=C A(\text { Common })}\end{array}\)
\(\begin{array}{l}{\angle B C A=\angle D A C(\text { Alternate interior angles, as } l \| m)} \\ { \triangle A B C = \Delta C D A(B y \text { ASA congruence rule) }}\end{array}\) 

Ans : \(\begin{array}{l}{\text { In } \triangle A P B \text { and } \triangle A Q B \text { . }} \\ {\angle A P B=\angle A Q B\left(E a c h 90^{\circ}\right)}\end{array}\)
\(\begin{array}{l}{\angle P A B=\angle Q A B(1 \text { is the angle bisector of } \angle A)} \\ {A B=A B(\text { Common })}\end{array}\)
\(\begin{array}{l}{\therefore \triangle \mathrm{APB} \cong \triangle \mathrm{AQB}(\mathrm{By} \text { AAS congruence rule) }} \\ {\therefore \mathrm{BP}=\mathrm{BQ}(\mathrm{By} \mathrm{CPCT})}\end{array}\)
Or, it can be said that B is equidistant from the arms of ∠ A.