D is a point on the side BC of a triangle ABC such that \(\angle\) ADC = \(\angle\) BAC. Show that \(CA^2 = CB.CD\)

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



In \(\triangle\) ADC and \(\triangle\) BAC,
\(\angle\) ADC = \(\angle\) BAC (Already given)
\(\angle\) ACD = \(\angle\) BCA (Common angles)
\(\triangle\) ADC ~ \(\triangle\) BAC (AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.
CA/CB = CD/CA
\(CA^2 = CB.CD\)
Hence, proved.