In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent.

Prove that the line NM produced bisects AB at P.

We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ AP² = PM x PN ……….(1)

From P, PB is the tangent and PMN is the secant for second circle.

∴ PB² = PM x PN ……… (2)

From (1) and (2), we have

⇒ AP² = PB²

⇒ AP = PB

Hence, proved that P is the midpoint of AB.