# 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 = $$\angle$$ BAC

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

$$\therefore$$  $$BC\over AC$$ = $$AC\over DC$$

$$\implies$$  $${CA}^2$$ = $$DC \times CB$$