Question

In the above figure, 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$$=$ $BC$ $+$ $DA.$

Answer 1

The figure shows that the tangents drawn from the exterior point to a circle are equal in length.

As $DR$ and $DS$ are tangents from exterior point $D$ so, $DR$ $=$ $DS$---- (1)

As $AP$ and $AS$ are tangents from exterior point $A$ so, $AP$ $=$ $AS$---- (2)

As $BP$ and $BQ$ are tangents from exterior point $B$ so, $BP$ $=$ $BQ$---- (3)

As $CR$ and $CQ$ are tangents from exterior point $C$ so, $CR$ $=$ $CQ$---- (4)

Adding the equation $1,2,3\&4$, we get

$DR+AP+BP+CR=DS+AS+BQ+CQ$

$(DR+CR)+(AP+BP)=(DS+AS)+(BQ+CQ)$

$CD+AB=DA+BC$

$AB+CD=BC+DA$

Hence proved.