Question

If ABCD is a trapezium in which AB parallel to CD parallel to EF, then prove that AE/ED=BF/FC

Answer 1

Attached Image (Created by me)

Draw AC intersecting EF at G.

In ΔCAB, GF || AB

$\bf\huge\frac{BF}{FC} = \frac{AG}{CG}$  By BPT ...(i)

In ΔADC, EG || DC

$\bf\huge\frac{AE}{ED} = \frac{BF}{FC}$ By BPT ...(ii)

From equations (i) and (ii),

$\bf\huge\frac{AE}{ED} = \frac{BF}{FC}$

$\bf\huge\boxed{\boxed{\boxed{\bf\:Proved}}}$