In the following Fig, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.
Since tangents drawn from an exterior point to a circle are equal in length,
AP = AS ….(1)
BP = BQ ….(2)
CR = CQ ….(3)
DR = DS ….(4)
Adding equations (1), (2), (3) and (4), we get
AP + BP + CR + DS = AS + BQ + CQ + DS
∴(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
∴AB + CD = AD + BC
∴AB + CD = BC + DA …..(proved)
Concept: Number of Tangents from a Point on a Circle
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