Question

$D$ is a point on the side $BC$ of $\triangle ABC$ such that $\angle ADC = \angle BAC$. Show that:
$AC^2 = BC \times DC$

Answer 1

We are given that $\angle ADC = \angle BAC$, which means triangle $ADC$ is similar to triangle $ABC$ by the AA (Angle-Angle) similarity criterion.

We can apply the following proportionality rule:

$\frac{AC}{BC} = \frac{DC}{AC}$

Now, cross-multiply to get:

$AC^2 = BC \times DC$

This is the required result.