D is a point on the side BC of ΔABC such that ∠ADC = ∠BAC. Prove that CA/CB = CD/CA or CA^2 = CB × CD

Question

D is a point on the side BC of ΔABC such that ∠ADC = ∠BAC. Prove that \(\frac{CA}{CD} = \frac{CB}{CA} \) or CA²  = CB × CD

Answer 1

In ΔABC and ΔDAC, we have

∠ADC = ∠BAC and ∠C = ∠C 

∴ By AA - axiom of similarity

ΔABC ~ ΔDAC 

⇒ \(\frac{AB}{DA}=\frac{BC}{AC} =\frac{AC}{DC}\) 

⇒ \(\frac{CB}{CA}=\frac{CA}{CD}\)