Solution

(i) It is given that in triangle $ABC,AB=AC$

$\Rightarrow \mathrm{\angle }ACB=\mathrm{\angle }ABC$ (Angles opposite to equal sides of a triangle are equal)

$\begin{array}{rlr}& \Rightarrow \frac{1}{2}\mathrm{\angle }ACB=\frac{1}{2}\mathrm{\angle }ABC& \Rightarrow \mathrm{\angle }OCB=\mathrm{\angle }OBC\end{array}$

$\Rightarrow \mathrm{OB}=\mathrm{OC}$ (Sides opposite to equal angles of a triangle are also equal)

(ii) In $\mathrm{△}OAB$ and $\mathrm{△}OAC$,

$\mathrm{AO}=\mathrm{AO}$ (Common)

$AB=AC($ Given $)$

$OB=OC$ (Proved above)

Therefore, $\mathrm{△}OAB\cong \mathrm{△}OAC$ (By SSS congruence rule)

$\Rightarrow \mathrm{\angle }BAO=\mathrm{\angle }CAO$ (CPCT)

$\Rightarrow \mathrm{AO}$ bisects $\mathrm{\angle }\mathrm{A}$.

Solution

In $\mathrm{△}ADC$ and $\mathrm{△}ADB$,

$AD=AD$ (Common)

$\therefore ADC=\therefore ADB($ Each 90) $CD=BD(AD$ is the perpendicular bisector of $BC)$

$\therefore \mathrm{△}ADC\therefore \mathrm{△}ADB$ (By SAS congruence rule)

$\therefore AB=AC(ByCPCT)$

Therefore, $ABC$ is an isosceles triangle in which $AB=AC$.

Solution

In $\mathrm{△}AEB$ and $\mathrm{△}AFC$,

$\therefore AEB$ and $AFC(.$ Each ${.90}^{\circ })A=$

$\therefore A$ (Common angle) $AB=AC$ (Given)

$\therefore \mathrm{△}AEB\therefore \mathrm{△}AFC$ (By AAS congruence rule) $\therefore BE=$ $CF$ (By CPCT)

Solution

(i) In $\mathrm{△}ABE$ and $\mathrm{△}ACF$,

$\therefore ABE$ and $ACF(.$ Each ${90}^{\circ }$ )

$\therefore A=A^{\prime }$ (Common angle)

$BE=CF$ (Given)

$\therefore \mathrm{△}ABE\therefore \mathrm{△}ACF$ (By AAS congruence rule)

(ii) It has already been proved that

$\mathrm{△}ABE\bullet \mathrm{△}ACF$

$\therefore AB=AC(ByCPCT)$

Solution

Let us join $AD$.

In $\mathrm{△}ABD$ and $\mathrm{△}ACD$,

$AB=AC$ (Given)

$BD=CD$ (Given)

$AD=AD$ (Common side)

$\therefore \mathrm{△}ABD\cong \mathrm{△}ACD$ (By SSS congruence rule)

$\therefore \therefore ABD=AACD(ByCPCT)$

Solution

In $\mathrm{△}ABC$,

$AB=AC$ (Given)

$\therefore \therefore ACB=\therefore ABC$ (Angles opposite to equal sides of a triangle are also equal)

In $\mathrm{△}ACD$,

$AC=AD$

$\therefore \therefore$ ADC $=:ACD$ (Angles opposite to equal sides of a triangle are also equal)

In $\mathrm{△}BCD$,

$\therefore ABC+BCD+ADC={180}^{\circ }$ (Angle sum property of a triangle)

$\therefore A\dot{CB}+ACB+ACD+\therefore ACD={180}^{\circ }$

$\therefore \therefore 2(ACB+ACD)={180}^{\circ }$

$\therefore \therefore 2(BCD)={180}^{\circ }$

$\therefore BCD={90}^{\circ }$

Solution

It is given that

$AB=AC$

$\therefore \dot{C}=B$ (Angles opposite to equal sides are also equal)

In $\mathrm{△}ABC$

$\therefore A+\dot{B}+C={180}^{\circ }$ (Angle sum property of a triangle)

$\therefore {90}^{\circ }+\stackrel{\therefore }{B}+C\stackrel{\therefore }{=}{180}^{\circ }$

$\therefore {90}^{\circ }+\stackrel{\therefore }{B}+B\stackrel{\therefore }{=}{180}^{\circ }$

$\therefore \therefore {90}^{\circ }$

$\therefore$ :

$B=450$

$B=C={45}^{\circ }$

Solution

Let us consider that $ABC$ is an equilateral triangle.

Therefore, $AB=BC=AC$

$AB=AC$

$\therefore C=B$ (Angles opposite to equal sides of a triangle are equal)

Also,

$AC=BC$

$\therefore B=A$ (Angles opposite to equal sides of a triangle are equal)

Therefore, we obtain $\therefore$ A

$=B\cdot C\therefore$

In $\mathrm{△}ABC$,

$\therefore A+B+C={180}^{\circ }$

$\therefore A^{\circ }+A+A+{180}^{\circ }$

$\therefore 3\dot{A}={180}^{\circ }$

$\therefore \dot{A}={60}^{\circ }$

$\therefore \dot{A}=B\stackrel{\mathrm{\Delta }}{=}C=\ddot{6}{0}^{\circ }$ Hence, in an equilateral triangle, all interior angles are of measure ${60}^{\circ }$.