#### Question

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.

#### Solution

From P, AP is the tangent and PMN is the secant for first circle.

∴ `AP^2 = PM × PN` …… (i)

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

∴ `PB^2 = PM × PN` ……..(ii)

From (i) and (ii)

`AP^2 = PB^2`

⇒ AP = PB

Therefore, P is the midpoint of AB.

Is there an error in this question or solution?

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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. Concept: Tangent Properties - If a Chord and a Tangent Intersect Externally, Then the Product of the Lengths of 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.

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