Answer:
Given, a quad. ABCD and a circle touches it?s all four sides at P, Q, R, and S respectively. To prove: \[AB+CD=BC+DA\] L.H.S. \[=AB+CD\] \[=AP+PB+CR+RD\] \[=AS+BQ+CQ+DS\] (Tangents from same external point are always equal) \[=(AS+SD)+(BQ+QC)\] \[=AD+BC\] = R.H.S. Hence Proved.
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