Solution

(i) In $\triangle A B D$ and $\triangle A C D$,

$ \begin{aligned} & A B=A C \ldots \text {…(since } \triangle A B C \text { is isosceles) } \\ & A D=A D \ldots \text {…(common side) } \\ & B D=D C \text {…(since } \triangle \mathrm{BDC} \text { is isosceles) } \\ & \triangle \mathrm{ABD} \cong \triangle \mathrm{ACD} \text {…..SSS test of congruence, } \\ & \therefore \angle \mathrm{BAD}=\angle \mathrm{CAD} \text { i.e. } \angle \mathrm{BAP}=\angle \mathrm{PAC} \text {…..[c.a.c.t]…..(i) } \end{aligned} $

(ii) In $\triangle A B P$ and $\triangle A C P$,

$ \begin{aligned} & A B=A C \ldots \text {…(since } \triangle A B C \text { is isosceles) } \\ & A P=A P \ldots \text {…ommon side) } \\ & \angle B A P=\angle P A C \text {….from (i) } \\ & \triangle A B P \cong \triangle A C P \text {…. SAS test of congruence } \\ & \therefore B P=P C \text {…[C.s.c.t]…..(ii) } \\ & \angle A P B=\angle A P C \text {….c.a.c.t. } \end{aligned} $

(iii) Since $\triangle A B D \cong \triangle A C D$

$ \angle \mathrm{BAD}=\angle \mathrm{CAD} \text {….from (i) } $

So, $A$ bisects $\angle A$

i.e. AP bisects $\angle A$…..(iii)

In $\triangle \mathrm{BDP}$ and $\triangle \mathrm{CDP}$,

DP $=$ DP …common side

$B P=P C$…from (ii)

$\mathrm{BD}=\mathrm{CD}$…(since $\triangle \mathrm{BDC}$ is isosceles)

$\triangle \mathrm{BDP} \cong \triangle \mathrm{CDP}$….SSS test of congruence

$\therefore \angle \mathrm{BDP}=\angle \mathrm{CDP}$….c.a.c.t.

$\therefore \mathrm{DP}$ bisects $\angle \mathrm{D}$

So, AP bisects $\angle \mathrm{D}$

From (iii) and (iv),

AP bisects $\angle \mathrm{A}$ as well as $\angle \mathrm{D}$.

(iv) We know that

$\angle A P B+\angle A P C=180^{\circ} \ldots$ (angles in linear pair)

Also, $\angle \mathrm{APB}=\angle \mathrm{APC}$… from (ii)

$\therefore \angle \mathrm{APB}=\angle \mathrm{APC}=\frac{180^{\circ}}{2}=90^{\circ}$

$\mathrm{BP}=\mathrm{PC}$ and $\angle \mathrm{APB}=\angle \mathrm{APC}=90^{\circ}$

Hence, $\mathrm{AP}$ is perpendicular bisector of $\mathrm{BC}$.

Solution

(i) In $\triangle B A D$ and $\triangle C A D$,

$\therefore A D B=\therefore A D C$ (Each $90^{\circ}$ as AD is an altitude)

$A B=A C$ (Given)

$A D=A D$ (Common)

$\therefore \quad \triangle BAD \therefore \triangle CAD$ (By RHS Congruence rule)

$\therefore \quad BD=CD(By CPCT)$

Hence, $A D$ bisects $B C$.

(ii) Also, by CPCT,

$\therefore B A D=C A D$ Hence, $A D$

bisects A. :

Solution

(i) In $\triangle A B C, A M$ is the median to $B C$.

$ \begin{aligned} & \therefore \quad BM={ }^{\frac{1}{2}} BC \\ & \therefore \quad Q N={ }^{\frac{1}{2}} QR \end{aligned} $

However, $BC=QR$

$\therefore{ }^{\frac{1}{2}} _{BC}{ }^{\frac{1}{2}} QR$

$\therefore \quad BM=QN \ldots$ (1)

In $\triangle A B M$ and $\triangle P Q N$, In $\triangle P Q R, P N$ is the median to $Q R$.

$ \begin{aligned} & A B=P Q \text{ (Given) } \\ & BM=QN[\text{ From equation (1)] } \\ & A M=P N(\text{ Given) } \\ & \therefore \quad \therefore \quad \triangle ABM \quad \triangle PQN \text{ (SSS congruence rule) } \\ & \therefore \quad \therefore \quad A B M=P Q N(B y C P C T) \\ & \therefore \quad \therefore \quad A B C=P Q R \ldots(2) \end{aligned} $

(ii) In $\triangle A B C$ and $\triangle P Q R$,

$A B=P Q$ (Given)

$\therefore A B C=\therefore P Q R$ [From equation (2)]

$BC=QR$ (Given)

$\therefore \triangle ABC \therefore \triangle PQR$ (By SAS congruence rule)

Solution

In $\triangle B E C$ and $\triangle C F B$,

$\therefore BEC=\therefore CFB(.$ Each $.90^{\circ})$

$BC=CB$ (Common)

$B E=C F$ (Given)

$\therefore \triangle BEC \quad \triangle CFB$ (By RHS congruency)

$\therefore B B^{\prime} C=C B ‘(B y C P C T)$

$\therefore A B=A C$ (Sides opposite to equal angles of a triangle are equal)

Hence, $\triangle ABC$ is isosceles.

Solution

In $\triangle A P B$ and $\triangle A P C$,

$\therefore APB=\therefore APC(.$ Each $.90^{\circ})$

$A B=A C($ Given $)$

$A P=A P($ Common $)$

$\therefore \triangle APB \quad \triangle APC$ (Using RHS congruence rule)

$\therefore B^{\prime \prime}=C$ (By using CPCT)