Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.
In ΔADC and ΔBAC,
∠ADC = ∠BAC (Already given)
∠ACD = ∠BCA (Common angles)
ΔADC ~ ΔBAC (AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
⇒ CA2 = CB.CD.
Hence, proved.Answered by Pragya Singh | 1 year ago