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

Given, D is a point on the side BC of a triangle ABC such that $$\angle$$ ADC = $$\angle$$ BAC.

In $$\triangle$$ ADC and $$\triangle$$ BAC,
$$\angle$$ ADC = $$\angle$$ BAC (Already given)
$$\angle$$ ACD = $$\angle$$ BCA (Common angles)
$$\triangle$$ ADC ~ $$\triangle$$ BAC (AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.
CA/CB = CD/CA
$$CA^2 = CB.CD$$
Hence, proved.