In given figure D is a point on the side BC of $△ A B C$ such that $∠ A D C = ∠ B A C$ . Prove that

$\frac{C A}{C D} = \frac{C B}{C A}$ or ${C A}^{2} = C B \times C D$ .

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Solution

In $△ A B C$ and $△ D A C$ , we have

$∠ A D C = ∠ B A C$ and $∠ C = ∠ C$

Therefore, by AA-criterion of similarity, we have

$△ A B C \sim △ D A C$

$\Rightarrow$ $\frac{A B}{D A} = \frac{B C}{A C} = \frac{A C}{D C}$

$\Rightarrow$ $\frac{C B}{C A} = \frac{C A}{C D}$ $[H e n c e p r o v e d]$

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