Prove it

Asked by  | 30th Jan, 2009, 06:27: PM

Expert Answer:

Consider the trapezium ABCD

with AB parallel to CD.

Let EF be the line parallel to AB  and CD, intersecting non parallel sides AD and BC in E and F respectively.

Join diagonal AC intersecting EF in O.

Consider triangle DAC,

We have ,

EO paralle to DC,

so by basic proportionality theorem, we get,

DE/EA=CO/OA...(1)

Similarly,

In triangle CAB,

we have

OF parallel to AB.

So,

 CO/OA=CF/FB ....(2)     ( by B.P.T.)

So from (1) and (2)

we get,

DE/EA=CF/FB

Hence proved.

Answered by  | 30th Jan, 2009, 10:36: PM

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