Question

In the above, a quadrilateral ABCD is drawn to circumscribe a circle, with center 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.

Answer 1

Hint: Use the property that “tangents drawn from a particular point to a circle are equal in length”. Form four relations using this property and add them in such an order to get AB and CD at one side and BC and DC at the other side.

Complete step by step answer:
We know that tangents drawn from a particular point to a circle are equal in length. Therefore,
Considering tangents AP and AS drawn from point A,
AP = AS – (1)
Considering tangents BP and BQ drawn from point B,
BP = BQ – (2)
Considering tangents CP and CQ drawn from point C,
CR = CQ – (3)
Considering tangents DR and DS drawn from point D,
DR = DS – (4)
Now adding equations (1), (2, (3) and (4), we get,
AP + BP + CR + DR = AS + BQ + CQ + DS
Grouping the term, we get,
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
\[\Rightarrow \] AB + CD = AD + BC
\[\Rightarrow \] AB + CD = BC + AD

Note: One may note that we have to group the terms such that we get AB and CD at one side and BC and AD at the other. If we do not take care then we may get confused and waste too much time in solving this problem.