Answers

2016-03-17T19:46:39+05:30
If parallelogram ABCD inscribed a circle
let p ,q,r,s are lie on AB,BC,CD,DA
since two tangents from an external points are equal in length
so PA=AS
PB=BQ
DR=DS
RC=CQ
adding THESE
PA+PB+DR+RC=AS+BQ+DS+CQ
AB+CD=AD+BC
(since ABCD is parallelogram )
in parallelogram opposite sides equal

so
AB+AB=AD+AD
2AB=2AD
AB=AD
since one pair of adjacent sides is equal
So ABCD is rhombus
proved
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