Prove that the centre of a circle touching two intersecting straight lines lies on the angle bisector of the lines.
Centre lies on each angle bisector.
Step 1: Let the two straight lines intersect at a point \(X\).
Step 2: Suppose a circle is drawn that touches both lines. Let the centre of the circle be \(O\), and let it touch the first line at point \(T_1\) and the second line at point \(T_2\).
Step 3: A basic property of circles is: the radius drawn to the point of contact with a tangent is perpendicular to that tangent.
So, \(OT_1 \perp \text{first line}\) and \(OT_2 \perp \text{second line}\).
Step 4: This means the perpendicular distance from the centre \(O\) to the first line is equal to \(OT_1 = r\) (radius), and the perpendicular distance from \(O\) to the second line is also equal to \(OT_2 = r\).
Step 5: Therefore, the centre \(O\) is equidistant from the two lines.
Step 6: In geometry, the set (locus) of all points that are equidistant from two intersecting lines is the pair of their angle bisectors.
Step 7: Hence, the centre of the circle must lie on the angle bisector of the given two lines.