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$ \mathrm{AD} $ is an altitude of an isosceles triangle $ \mathrm{ABC} $ in which $ \mathrm{AB}=\mathrm{AC} $. Show that
(i) ADbisects BC
(ii) $ \mathrm{AD} $ bisects $ \angle \mathrm{A} $.
Given:
$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB=AC$.
To do:
We have to show that:
(i) $AD$ bisects $BC$
(ii) $AD$ bisects $\angle A$.
Solution:
Let us consider $\triangle ABD$ and $ACD$
Given, that $AD$ is the altitude of $\triangle ABD$ and $ACD$
We get, $\angle ADB=\angle ADC=90^o$
We have, $AB=AC$
Since $AD$ is the common side we get,
$AD=DA$
Therefore,
We know that According to the RHS rule if the hypotenuse and one side of a right-angled triangle are equal to the corresponding hypotenuse and one side of another right-angled triangle; then both the right-angled triangle are said to be congruent.
Therefore,
$\triangle ABD \cong \triangle ACD$
We also know that,
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding sides must be equal.
Therefore,
$BD=CD$
Hence, $AD$ bisects $BC$.
(ii) Again from CPCT,
We also know
From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles must be equal.
Hence,
$AD$ bisects $\angle A$.
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