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

Show that $C{A}^2=CB.CD$

Given that

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

To Prove $C{A}^2=CB.CD$

Proof

From$\Delta ADC$ and $\Delta BAC,$

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

$\angle ACD=\angle BCA$ (Common angles)

$\therefore \Delta ADC~\Delta BAC\mathit{}$ (From AA similarity criterion)

We know that the corresponding sides of similar triangles are in proportion.

$\therefore \frac{CA}{CB}=\frac{CD}{CA}$

Hence

$\Rightarrow \mathit{C}{\mathit{A}}^{\mathbf{2}}\mathbf{=}\mathbf{}\mathit{C}\mathit{B}\mathbf{.}\mathit{C}\mathit{D}\mathbf{.}$

Hence, proved.