Question

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Answer 1

Assume that $O$ touches $l_1$ and $l_2$ at $M$ and $N$ , then we get as ,

$\Rightarrow OM=ON$ ( As it is the radius of the circle )

Therefore ,From the centre ${}^{\prime }{O}^{\prime }$ of the circle , it has equal distance from $l_1$ and $l_2$

Now , In $△OPMandOPN$

$\Rightarrow OM=ON$ ( Radius of the circle )

$\Rightarrow \angle OMP=\angle ONP$ ( As , Radius is perpendicular to its tangent )

$\Rightarrow OP=OP$ ( Common sides )

Therefore ,$△OPM\cong △OPN$ ( S S S congruence rule )

By C.P.C.T ,

$\Rightarrow \angle MPO=\angle NPO$

So ,$l$ bisects $\angle MPN$.

Therefore , $O$ lies on the bisector of the angle between$l_1,l_2$

Hence , we prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines