Let the lines be l1 and l2.
Assume that O touches l₁ and l₂ at M and N,
We get,
OM = ON (Radius of the circle)
Therefore,
From the centre ”O” of the circle, it has equal distance from l₁ & l₂.
In Δ OPM & OPN,
OM = ON (Radius of the circle)
∠OMP = ∠ONP (As, Radius is perpendicular to its tangent)
OP = OP (Common sides)
Therefore,
Δ OPM = ΔOPN (SSS congruence rule)
By C.P.C.T,
∠MPO = ∠NPO
So, l bisects ∠MPN.
Therefore, O lies on the bisector of the angle between l₁ & l₂ .
Hence, we prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.