Question

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB+CD=AD+BC.
or
A circle touches all the four sides of a quadrilateral ABCD. Prove that AB+CD=BC+DA.

Answer 1

Let the circle touches the sides AB,BC,CD and DA of a squareABCD at P,Q,R and S respectively. Since, the tangents drawn from an external point to a circle are equal in length.
`:." "AP=AS" "...(1)`
`BP=BQ" "...(2)`
`CR=CQ" "...(3)`
and`" "DR=DS" "...(4)`
Adding (1), (2), (3) and (4), we get
`ubrace(AP+BP)+ubrace(CR+DR)=AS+BQ+CQ+DS`
`implies" "AB+CD=(AS+DS)+(BQ+CQ)`
`implies" "AB+CD=AD+BC" "` Hence Proved.