Question

In the given figure ∠CEF = ∠CFE. F is the midpoint of DC. Prove that `(AB)/(BD) = (AE)/(FD)`

Answer 1

Draw DG || BE

In ΔABE, `(AB)/(BD) = (AE)/(GE)`  ...[BPT]

CF = FD    [F is the midpoint of DC] ...(i)

In ΔCDG, `(DF)/(CF) = (GE)/(CE)` = 1  ...[Mid point theorem]

GE = CE  ...(ii)

∠CEF = ∠CFE  ...[Given]

CF = CE  [Sides opposite to equal angles]  ...(iii)

From (ii) and (iii) CF = GE  ...(iv)

From (i) and (iv) GE = FD

∴ `(AB)/(BD) = (AE)/(GE) \implies (AB)/(BD) = (AE)/(FD)`