Question
$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle$$ADC =$ $\angle$$BAC$. Show that $CA^2= CB\cdot$$CD$.
✓
Answer
Given:$\triangle\mathrm{ABC}$
where $\angle$$ADC =$ $\angle$$BAC$
To Prove :$ CA^2= CB.CD$
Proof: In $\triangle$$ABC$ and $\triangle$$DAC$, we have
$\angle$$ADC =$ $\angle$$BAC$ and $\angle$$C =$ $\angle$$C$
Therefore, by $AA$-criterion of similarity, we obtain
$\Delta A B C \sim \Delta D A C$
$\Rightarrow \quad \frac { A B } { D A } = \frac { B C } { A C } = \frac { A C } { D C }$
$\Rightarrow \quad \frac { C B } { C A } = \frac { C A } { C D }$
$\Rightarrow CA^2= CB.CD$
where $\angle$$ADC =$ $\angle$$BAC$
To Prove :$ CA^2= CB.CD$
Proof: In $\triangle$$ABC$ and $\triangle$$DAC$, we have
$\angle$$ADC =$ $\angle$$BAC$ and $\angle$$C =$ $\angle$$C$
Therefore, by $AA$-criterion of similarity, we obtain
$\Delta A B C \sim \Delta D A C$
$\Rightarrow \quad \frac { A B } { D A } = \frac { B C } { A C } = \frac { A C } { D C }$
$\Rightarrow \quad \frac { C B } { C A } = \frac { C A } { C D }$
$\Rightarrow CA^2= CB.CD$
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