Question

in the fig. angle BED =angle BDE and E is the moddle point of BC . Prove that AC/CF =AD/BE

Answer 1

See diagram...  There is a mistake in the ratio to be proved.... It is 
           AF / CF = AD / BE.

Let ∠BDE = ∠BED = x°.   
Hence, in ΔBDE, we have BD = BE = BC/2.

Now Draw CG parallel to AB.   
∠CEG = x°  (vertical angle). 

 ΔBED and Δ CEG are similar as   CE || BE, CG || BD, EG || ED.
Since BE = EC, They are congruent.  
Hence, ∠CEG = ∠CGE = x°   and    ∠ECG = ∠B.

Now for the ΔABC, ∠FCB is external angle =>    ∠FCB = ∠A+∠B
Hence,   ∠GCF = ∠A.
Further, ∠CGF = 180° - ∠CGE = 180° - x°.

In ΔCGF , ∠F = 180° - ∠A - (180° - x°) = x° - ∠A

Now compare ΔADF  and  Δ CGF:  Since all corresponding angles are same they are similar triangles.

=>  AF/AD = CF/CG = CF / CE = CF / BE
=>  AF / CF = AD / BE