Solution

(i) In $\mathrm{△}ABD$ and $\mathrm{△}ACD$,

$\begin{array}{rl}& AB=AC\dots \text{…(since }\mathrm{△}ABC\text{ is isosceles) }\\ & AD=AD\dots \text{…(common side) }\\ & BD=DC\text{…(since }\mathrm{△}\mathrm{BDC}\text{ is isosceles) }\\ & \mathrm{△}\mathrm{ABD}\cong \mathrm{△}\mathrm{ACD}\text{…..SSS test of congruence, }\\ & \therefore \mathrm{\angle }\mathrm{BAD}=\mathrm{\angle }\mathrm{CAD}\text{ i.e. }\mathrm{\angle }\mathrm{BAP}=\mathrm{\angle }\mathrm{PAC}\text{…..[c.a.c.t]…..(i) }\end{array}$

(ii) In $\mathrm{△}ABP$ and $\mathrm{△}ACP$,

$\begin{array}{rl}& AB=AC\dots \text{…(since }\mathrm{△}ABC\text{ is isosceles) }\\ & AP=AP\dots \text{…ommon side) }\\ & \mathrm{\angle }BAP=\mathrm{\angle }PAC\text{….from (i) }\\ & \mathrm{△}ABP\cong \mathrm{△}ACP\text{…. SAS test of congruence }\\ & \therefore BP=PC\text{…[C.s.c.t]…..(ii) }\\ & \mathrm{\angle }APB=\mathrm{\angle }APC\text{….c.a.c.t. }\end{array}$

(iii) Since $\mathrm{△}ABD\cong \mathrm{△}ACD$

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

So, $A$ bisects $\mathrm{\angle }A$

i.e. AP bisects $\mathrm{\angle }A$…..(iii)

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

DP $=$ DP …common side

$BP=PC$…from (ii)

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

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

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

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

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

From (iii) and (iv),

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

(iv) We know that

$\mathrm{\angle }APB+\mathrm{\angle }APC={180}^{\circ }\dots$ (angles in linear pair)

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

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

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

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

Solution

(i) In $\mathrm{△}BAD$ and $\mathrm{△}CAD$,

$\therefore ADB=\therefore ADC$ (Each ${90}^{\circ }$ as AD is an altitude)

$AB=AC$ (Given)

$AD=AD$ (Common)

$\therefore \mathrm{△}BAD\therefore \mathrm{△}CAD$ (By RHS Congruence rule)

$\therefore BD=CD(ByCPCT)$

Hence, $AD$ bisects $BC$.

(ii) Also, by CPCT,

$\therefore BAD=CAD$ Hence, $AD$

bisects A. :

Solution

(i) In $\mathrm{△}ABC,AM$ is the median to $BC$.

$\begin{array}{rl}& \therefore BM={}^{\frac{1}{2}}BC\\ & \therefore QN={}^{\frac{1}{2}}QR\end{array}$

However, $BC=QR$

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

$\therefore BM=QN\dots$ (1)

In $\mathrm{△}ABM$ and $\mathrm{△}PQN$, In $\mathrm{△}PQR,PN$ is the median to $QR$.

$\begin{array}{rl}& AB=PQ\text{ (Given) }\\ & BM=QN[\text{ From equation (1)] }\\ & AM=PN(\text{ Given) }\\ & \therefore \therefore \mathrm{△}ABM\mathrm{△}PQN\text{ (SSS congruence rule) }\\ & \therefore \therefore ABM=PQN(ByCPCT)\\ & \therefore \therefore ABC=PQR\dots (2)\end{array}$

(ii) In $\mathrm{△}ABC$ and $\mathrm{△}PQR$,

$AB=PQ$ (Given)

$\therefore ABC=\therefore PQR$ [From equation (2)]

$BC=QR$ (Given)

$\therefore \mathrm{△}ABC\therefore \mathrm{△}PQR$ (By SAS congruence rule)

Solution

In $\mathrm{△}BEC$ and $\mathrm{△}CFB$,

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

$BC=CB$ (Common)

$BE=CF$ (Given)

$\therefore \mathrm{△}BEC\mathrm{△}CFB$ (By RHS congruency)

$\therefore B{B}^{\prime }C=CB‘(ByCPCT)$

$\therefore AB=AC$ (Sides opposite to equal angles of a triangle are equal)

Hence, $\mathrm{△}ABC$ is isosceles.

Solution

In $\mathrm{△}APB$ and $\mathrm{△}APC$,

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

$AB=AC($ Given $)$

$AP=AP($ Common $)$

$\therefore \mathrm{△}APB\mathrm{△}APC$ (Using RHS congruence rule)

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