Question

Suppose ABC is an isosceles triangle such that AB = AC and AD is the altitude from A on BC. Prove that (i) AD bisects $\angle A$, (ii) AD bisects BC.

Answer 1

We have to show that
$\angle BAD=\angle CAD$ and BD = DC.
In right triangles ADB and ADC, we have
AB = AC (given);
AD = AD (common side).
So by RHS congruency of triangles, we have $△ABD\cong △ACD$. Hence, $\angle BAD=\angle CAD$ and $BD=DC$.