$A D$ is an altitude of an isosceles triangles $A B C$ in which $A B=A C$. Show that

(i) $A D$ bisects $B C$

(ii) $\mathrm{AD}$ bisects $\angle \mathrm{A}$.
Solution:

"

(i) $\ln \triangle \mathrm{BAD}$ and $\triangle \mathrm{CAD}$,

$\angle A D B=\angle A D C\left(\right.$ Each $90^{\circ}$ as $A D$ is an altitude $)$

$A B=A C$ (Given)

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

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

$\Rightarrow \mathrm{BD}=\mathrm{CD}(\mathrm{By} \mathrm{CPCT})$

Hence, AD bisects BC.

(ii) Also, by CPCT,

$\angle \mathrm{BAD}=\angle \mathrm{CAD}$

Hence, AD bisects $\angle \mathrm{A}$.
Editor