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Question
In triangle ABC point D is on side BC (B−D−C) such that ∠BAC = ∠ADC then prove that CA2 = CB × CD
Solution
In ΔBAC and ΔADC,
∠BAC ≅ ∠ADC ......[Given]
∠BCA ≅ ∠ACD ......[Common angle]
∴ ΔBAC ∼ ΔADC .....[AA test of similarity]
∴ `"CA"/"CD" = "CB"/"CA"` ......[Corresponding sides of similar triangles]
∴ CA × CA = CB × CD
∴ CA2 = CB × CD
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