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Question
In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD
Solution
Given: ∠BAC = ∠ADC
To prove: CA2 = CB × CD
Proof: In ∆ABC and ∆DAC
∠BAC = ∠ADC (Given)
∠C = ∠C (Common)
By AA test of similarity
∆ABC ∼ ∆DAC
\[\therefore \frac{BC}{AC} = \frac{AC}{DC} \left( \text{ Corresponding sides are proportional } \right)\]
\[ \Rightarrow {AC}^2 = BC \times DC\]
Hence proved.
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