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Asked by ashwati | 08 Mar, 2009, 06:12: PM

Construction : Draw a line PR parallel to CD , R lies on AD.

From corollary to mid point theoram in triangle ADC , as PR|| CD, P being the mid point , R is the mid point of AD.

As CD||AB, we have PR|| CD..Consider the triangle ABC, we have R as mid point of AD and similar as above it meets BD at midpoint i.e Q. Thus PQ||AB.

Now applying midpoint theoram in triangle ADC, RP=1/2 CD

and applying the midpoint theoram in ABD,RQ=1/2 AB

So we have , PQ=RP-RQ=1/2 (CD-AB)

Note: Above solution assumes that CD is the  larger side.

Answered by | 08 Mar, 2009, 09:06: PM

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