Question

If the sides of a quadrilateral touch a circles, prove that the sum of opposite sides is equal to the sum of the other pairs.

Answer 1

Now

$\because$ The length of tangents drawn from an external point to a circle are equal

$\therefore \Rightarrow DR=DS....\left(1\right)$

$\Rightarrow CR=CQ....\left(2\right)$

$\Rightarrow BQ=BP....\left(3\right)$

$\Rightarrow AP=AS....\left(4\right)$

Now,

Adding the four equations $\left(1\right)+\left(2\right)+\left(3\right)+\left(4\right)$

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

$\Rightarrow (DR+CR)+(AP+BP)=(DB+AS)+(CQ+BQ)$

Now,

$\because DR+CR=DC,AP+BP=AB$

$DS+AS=AD,CQ+BQ=BC$

$\therefore DC+AB=DA+CB$ Hence proved