D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$, prove that $CA^2 = CB \cdot CD$.

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Step 1: Triangle ABC with point D on side BC, angle ADC equals angle BAC
Step 2: In $	riangle CBA$ and $	riangle CDA$: $\angle C = \angle C$ (common)
Step 3: $\angle BAC = \angle ADC$ (given)
Step 4: $	herefore 	riangle CBA \sim 	riangle CAD$ (AA similarity)
Step 5: $	herefore \frac{CB}{CA} = \frac{CA}{CD} \Rightarrow CA^2 = CB \cdot CD$. Hence proved.

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