In the given figure, ∠AEF=∠AFE and E is the mid point of . . .

Question: In the given figure, ∠AEF=∠AFE and E is the mid point of CA. Prove that BD/CD = BF/CE.

Doubt by Suraj 

Solution : 

Given : ∠AEF=∠AFE i.e. ∠1=∠2

E is the mid point of CA i.e AE=CE

To Prove : BD/CD = BF/CE

Construction : Take point G on AB such that CG||DF

Proof : 

∠1=∠2 (Given) 
AE=AF (Sides opposite to equal angles of a triangle are equal) — (1) 

Also, 
AE=CE (Given) — (2) 

In ΔAGC
E is the mid point of AC (Given)
CG||EF [∵CG||DF]
⇒F is the mid point of GA
(By Mid Point Theorem)
 AF=GF — (3)

In ΔBDF
CG||DF (By Const.) 
BC/CD = BG/GF (By BPT)
Adding 1 both sides
[BC/CD] + 1 = [BG/GF] +1
[BC+CD]/CD = [BG+GF]/GF
BD/CD = BF/GF
BD/CD = BF/CE

[Using equation (1), (2) and (3)
GF=AF=AE=CE]

Hence Proved.