Prove that the centre of a …

Let l₁ and l₂ be two intersecting lines.

Suppose a circle with centre O touches the lines
l₁ and l₂ at M and N respectively.
{tex}{/tex} Therefore,OM = ON
{tex}{/tex}Therefore, O is equidistant from l₁ and l₂.
Consider  {tex}\Delta{/tex}OPM and {tex}\Delta{/tex}OPN,
{tex}\angle OMP = \angle ONP{/tex} ....(radius is perpendicular to the tangent)
OP = OP ...(Common side)
OM = ON ...(radii of the same circle)
{tex}\Rightarrow \Delta OPM \cong \Delta OPN{/tex} ...(RHS congruence criterion)
{tex}\Rightarrow \angle MPO = \angle NPO{/tex} ...(CPCT)
{tex}\Rightarrow{/tex} l bisects {tex}\angle MPN{/tex}.
{tex}\Rightarrow{/tex} O lies on the bisector of the angles between
l₁ and l₂, that is, O lies on l.
Therefore, the centre of the circle touching two intersecting lines lies on the angles bisector of the two lines.