Question

In the figure, in ΔABC, point D on side BC is such that,∠BAC=∠ADC. Prove that, CA2=CB*CD

Answer 1

In ΔACD & ΔBCA, 

 ∠BAC=∠ADC   (Given) 

∠ACD=∠BCA  (Same angle) 

=> ΔACD ≈ ΔBCA (AA) 

Since, both triangles are similar ,since two corresponding angles are equal .

So, we used similarity criterion AA (angle angle) , to enhance similarity of triangle ACD and BCA .

So,Coresponding sides of similar triangles are proportional .

 $\frac{AC}{BC} =  \frac{CD}{C A}  \\ \\ =\ \textgreater \  AC^{2} = BC *CD$ 

Hence Proved , $AC^2=BC*CD \\ CA^2=CB*CD$

Answer 2

Solution :-

please check attachment