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D is a point on the side BC of a triangl...

D is a point on the side BC of a triangle ABC such that `/_A D C=/_B A C`. Show that `C A^2=C B.C D`.

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To solve the problem, we need to show that \( CA^2 = CB \cdot CD \) given that \( \angle ADC = \angle BAC \). ### Step-by-Step Solution: 1. **Identify the triangles**: We have triangle \( ABC \) and point \( D \) on side \( BC \). We know that \( \angle ADC = \angle BAC \). 2. **Establish similarity**: Since \( \angle ADC = \angle BAC \) and \( \angle ACD \) is common to both triangles \( ADC \) and \( ABC \), we can conclude that: \[ ...
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