Question

In the given figure, line segment DF intersect the side AC of a triangle ΔABC at the
point E such that E is the mid–point of CA and ∠AEF = ∠AFE. Prove that:
BD/CD= BF/CE

[Hint: Take point G on AB such that CG || DF.]

Answer 1

Given that E is the mid point of AC ,then
CE=AE...............................1
also given that ,
angle AEF=angle AFE................2
then, AE=AF........................3
by eq1 and eq2 we get ,
CE=AE=AF........................4
now draw CG parallel DF
we know that corresponding angles are equal,
therefore,angleAEF=angle ACG........5
also, angle AFE=angle AGC...............6
by eq2,5,6, we get
angleAEF=AFE=ACG=angleAGC
Then,AC=AG.......................7. as sides opposite to equal angles
AE+CE=AF+GF
By eq4,7 we get,
AE=AF=GF=CE
CE=AF=GF.............................8
By Thales theorem we get,
BC/CD=BG/GF
By adding 1 on both sides we get,
BC/CD+1=BG/GF+1
(BC+CD)/CD=(BG+GF)/GF
BD/CD=BF/GF
by eq 8 we get
BD/CD=BF/CE
Hence proved

Answer 2

Answer:

all these answers are correct.

Step-by-step explanation:

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