In the given figure, D is a point on the side BC of $ riangle ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CD \cdot CB$.

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Question

In the given figure, D is a point on the side BC of $	riangle ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CD \cdot CB$.

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Accepted Answer

Step 1: In $	riangle ACD$ and $	riangle BCA$:
$\angle ACD = \angle BCA$ (common)
$\angle ADC = \angle BAC$ (given)
Step 2: $	herefore 	riangle ACD \sim 	riangle BCA$ (AA similarity)
So, $\frac{CA}{CB} = \frac{CD}{CA}$
$\Rightarrow CA^2 = CD \cdot CB$

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