In the given figure (ii) given below, $∠ A D C = ∠ B A C$ . Prove that $C A^{2} = D C \times B C$ .

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Solution

From the figure, $∠ A D C = ∠ B A C$
So, we have to prove that, $C A^{2} = D C \times B C$
In $△ A B C$ and $△ A D C$
$∠ C = ∠ C$ (common angle for both triangle)
$∠ B A C = ∠ A D C$ (from the question)
$△ A B C \sim △ A D C$
Therefore, $C A / D C = B C / C A$
We know that, corresponding sides are proportional,
Therefore, $C A_{2} = D C \times B C$ .

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D is a point on the side BC of $Δ A B C$ such that $∠ A D C = ∠ B A C$ . Prove that $\frac{C B}{C A}$ = $\frac{C A}{D C}$ or $C A^{2}$ = $C B \times C D$ .

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In the given figure (ii) given below, ∠ADC=∠BAC. Prove that CA2=DC×BC.

From the figure, ∠ADC=∠BACSo, we have to prove that, CA2=DC×BCIn △ABC and △ADC∠C=∠C (common angle for both triangle)∠BAC=∠ADC (from the question)△ABC∼△ADCTherefore, CA/DC=BC/CAWe know that, corresponding sides are proportional,Therefore, CA2=DC×BC.

In the given figure (ii) given below, $∠ A D C = ∠ B A C$ . Prove that $C A^{2} = D C \times B C$ .