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In the figure, in ΔABC, point D on side BC such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD

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Sol.     In ΔABC and ΔDAC
            ∠BAC ≅ ∠ADC                      [Given]
            ∠ACB ≅ ∠DCA                      [Common angle]
∴          ΔABC ~ ΔDAC                      [AA test of similarity]

∴          \({AB\over AD} = {BC\over AC} ={ AC\over CD}\)                   [Corresponding sides of similar triangles are proportional]

∴          \({BC\over AC} = {AC\over CD}\)

∴          AC2 = BC × CD                    [Cross multiplying]

Similarity August 23 , 2018 0 Comments 243 views