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