Question

$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC=\angle BAC$. Show that $C{A}^2=CB.CD$.

Answer 1

In $△ADC$ and $△BAC$

$\angle ADC=\angle BAC$ (Given)

$\angle C$ is Common

$\therefore$ by AA Criterion of Similarity, $△ADC$ $\sim$ $△BAC$

$\Rightarrow \frac{AD}{BA}=\frac{DC}{AC}=\frac{AC}{BC}$

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

$\therefore$ $C{A}^2=CB.CD$