D is a point on the side BC of \(\triangle\) ABC such that \(\angle\) ADC and \(\angle\) BAC are equal. Prove that \({CA}^2\) = \(DC \times CB\).

Solution :

Given : D is a point on the side BC of a triangle ABC, such that \(\angle\) ADC = \(\angle\) BACtriangles

To Prove : \({CA}^2\) = \(DC \times CB\)

Proof : In triangles ABC and DAC,

\(\angle\) BAC = \(\angle\) ADC       (given)

\(\angle\) C = \(\angle\) C                (common)

\(\angle\) ABC = \(\angle\) DAC        (third angles of the triangles)

\(\therefore\) \(\triangle\)s ABC and DAC are equiangular and hence, similar

\(\therefore\)  \(BC\over AC\) = \(AC\over DC\)

\(\implies\)  \({CA}^2\) = \(DC \times CB\)

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