Question

In the given figure, if ABCD is a trapezium in which AB || CD || EF, then prove that `(AE)/(ED) = (BF)/(FC)`.

Answer 1

Given:- AB || CD || EF

To prove:- `(AE)/(ED) = (BF)/(FC)`

Constant:- Join BD to intersects EF at G.

Proof:- In ΔABD

EG || AB (EF || AB)

`(AE)/(ED) = (BG)/(GD)`  (By BPT)  ...(1)

In ΔDBC

GF || CD (EF || CD)

`(BF)/(FC) = (BG)/(GD)`  (By BPT)  ...(2)

From (1) and (2)

`(AE)/(ED) = (BF)/(FC)`

Hence Proved