Question

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

Answer 1

Data: D is a point on the side BC of a triangle $△ABC$ such that $\angle ADC=\angle BAC.$

To Prove: $C{A}^2=CB\times CD$
Let $\angle ADC=\angle BAC={100}^{\circ }$
In $△ABC,$ if $\angle B={50}^{\circ },$ then $\angle C={30}^{\circ }$
In $△ADC,$ if $\angle C={30}^{\circ },$ then $\angle DAC={50}^{\circ }$
In $△BCP,\angle A={100}^{\circ },\angle B={50}^{\circ },\angle C={30}^{\circ }$
In $△ADC,\angle ADC=100,\angle DAC={50}^{\circ },\angle ACD={30}^{\circ }$
Similarity criterion of $△$ is A.A.A.
$\therefore$ In $△ABC$ and $△ADC,$
$\frac{CA}{BC}=\frac{DC}{CA}$
$\therefore CA\times CA=BC\times DC$
$\therefore C{A}^2=BC\times DC.$