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

In ΔADC and ΔBAC
∠ADC = ∠BAC (Given)
∠C is common
∴ by AA Criterion of Similarity, ΔADC ∼ ΔBAC
⇒ AD/BA = DC/AC = AC/BC
⇒ DC/AC = AC/BC
∴ CA² = CB.CD