Question

in the given figure angle BED is equal to Angle BDE and E is the midpoint of BC.
prove that AF/CF=AD/BE

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Answer 1

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