ICSE Class 10 Answered

As E is the mid point of AC, then

CE = AE    … (i)

Also, angle AEF = angle AFE    … (ii)

So, AE = AF    … (iii)

From (i) and (ii), we get

CE = AE = AF    … (iv)

Now draw CG || DF

As the corresponding angles are equal, so we have

angleAEF = angle ACG     … (v)

Also, angle AFE = angle AGC      … (vi)

From equations (ii), (v) and (vi), we get

angleAEF = angleAFE = angleACG = angleAGC

Therefore, AC = AG  … (Sides opposite to equal angles are equal)   (vii)

AE + CE = AF + GF

By equation (iv) and (vii) we get,

AE = AF = GF = CE

CE = AF = GF    … (viii)

By Thales theorem we get,

BC/CD = BG/GF

Adding 1 on both sides, we get

BC/CD + 1 = BG/GF + 1

(BC+CD)/CD = (BG+GF)/GF

BD/CD = BF/GF

From (viii), we get

BD/CD = BF/CE

Hence proved