$A D$ is an altitude of an isosceles triangles $A B C$ in which $A B=A C$. Show that<br/><br/>(i) $A D$ bisects $B C$<br/><br/>(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 $\angle \mathrm{A}$.
(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 $\angle \mathrm{A}$.
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