Question

ABCD is a rectangle and E is the point of BC.DB andAE intersect at F. prove that DF = 2FB and AF = 2FE

Answer 1

Solution:-
Given ABCD is a rectangle and BE = EC
To prove DF = 2FB and AF = 2FE
Proof : 
In the triangles BFE and DFA
∠ BFE = ∠ DFA    (vertically opposite angles)
∠ FBE = ∠ FDA   (alternate interior angles since AD || BC)
∠ FEB = ∠ FAD   (alternate interior angles)
Therefore by AAA similarity ; Δ BFE is congruent to Δ DFA
In similar triangles, the ratio of the corresponding sides are equal.
Therefore,
FB/DF = FE/AF = BE/AD .....(1)
BE = 1/2BC  (Since E is the mid point of BC)
BE = 1/2AD  {Since AD = BC  (Opposite sides of rectangle are equal)}
BE/AD = 1/2 ....(2)
From equation (1) and (2):
FB/DF = FE/AF = 1/2
⇒ DF = 2FB and AF = 2FE

Answer 2

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