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Question

A quadrilateral ABCD is drawn to circumscribe a circle (see Fig.) . Prove that AB+CD=AD+BC


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Solution

Circle properties:

Let theP,Q,R,Sbe point of contacts for tangent AB,BC,CD,DA respectively from the figure.

We know that the length of two tangents from one point is equal.

So,

AP=AS..(1)BP=BQ..(2)CR=CQ..(3)DR=DS..(4)

By, adding),(1),(2),(3) and(4) , we get

AP+BP+CR+DR=AS+BQ+CQ+DS(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)(OnRearranging)AB+CD=AD+BC

Hence, AB+CD=AD+BC proved.


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