Prove it
Asked by | 30th Jan, 2009, 06:27: PM
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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