Question

In given figure, $D$ is a point on the side $BC$ of $△ABC$ such that $\angle BAC=\angle ADC$ prove that
${CA}^2=CB\times CD$

Answer 1

Given:- $△ABC$ in which $\angle ADC=\angle BAAC$

To prove:- ${CA}^2=CB\cdot CD$

Proof:-

In $△ABC$ and $△ADC$

$\angle ACB=\angle ACD\left(\text{Common}\right)$

$\angle BAC=\angle ADC\left(\text{Given}\right)$

By AA similarity,

$△ABC\sim △ADC$

As we know that if triangles are similar then their corresponding sides are proportional.

$\therefore \frac{BA}{AD}=\frac{AC}{CD}=\frac{BC}{AC}$

Therefore,

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

$\Rightarrow {AC}^2=BC\cdot CD$

Hence proved.