Question

In a given figure if abcd is a trapezium in which AB || CD || EF , then prove that
AE\ED = BF\ FC

Answer 1

hope this helps you .....

Answer 2

Answer:

Given:

ABCD is a trapezium, where AB║DC.

E & F are points non-parallel sides AD & BC such that EF║AB.

To prove:

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

Proof:

Given, AB║DC & EF║AD

So, EF║DC  (∵Lines which are parallel to same line are parallel to each other)

Joining A & C

Let AC intersect EF at point G

Now, in ΔADC

EG║DC  (Because EF║DC)

So, $\frac{AE}{ED}=\frac{AG}{GC}$   ............(1)    (∵Line drawn parallel to one side of triangle, intersects the other two sides in distinct points. Then it divides the other 2 side in same ration)

Similarly, in ΔCAB

AB║GF  (∵ EF║AB)

So, $\frac{AG}{GC} =\frac{BF}{FC}$   ..........(2)     (∵ Line drawn parallel to one side of triangle, intersects the other two sides in distinct points. Then, it divides the other 2 side in same ratio)

From (1) & (2)

$\frac{AE}{ED}=\frac{AG}{GC}=\frac{BF}{FC}$

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

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