In figure, line segment DF intersects the side AC of a `DeltaABC` at the point E such that E is the mid-point of CA and `angleAEF-angleAFE`. Prove that `(BD)/(CD)=(BF)/(CE)`

Explanation

Given `DeltaABC`, E is the mid-point of CA and `angleAEF=angleAFE`
To prove `(BD)/(CD)=(BF)/(CE)`
Construction Take a point G on AB such that CG`abs()`EF.
Proof Since, E is the mid point of CA.
`therefore`CE=AE…(i)
In `DeltaACG,CGabs()Efand E` is the mid point of CA....(ii)
So, CE=GF...(ii)
[by mid point theorem]
Now in `DeltaBCG and DeltaBDF, CGabs()EF`
`therefore (BC)/(CD)=(BG)/(GF)` [ by basic proportionality theorem]
`rArr (BC)/(CD)=(BF-GF)/(GF)rArr(BC)/(CD)-1`
`rArr(BC)/(CD)+1=(BF)/(CE)` [from Eq. (ii)]
`rArr (BC+CD)/(CD)=(BF)/(CE)rArr(BD)/(CD)=(BF)/(CE)` Hence proved.