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