Question

D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC, Prove that CA² = CB.CD.

Answer 1

In ∆ADC and ∆BAC,

∠ADC = ∠BAC   [Given that]

∠C = ∠C   [Common]

\(\therefore \) By AA Criteria of similarity,

∆ADC \(\sim\) ∆BAC

\(\frac {AD}{BA} = \frac{DC}{AC} = \frac {AC}{BC}\)

\( \frac{DC}{AC} = \frac {AC}{BC}\)

\(CA^2 = CB.CD\)

Hence proved.