In two concentric

We are given two concentric circles C₁ and C₂ with centre O and a chord AB of the larger circle C₁ which touches the smaller circle C₂ at the point P (see Fig. 10.8).

We need to prove that AP = BP.

Let us join OP. Then, AB is a tangent to C₂ at P and OP is its radius. Therefore,

OP⟂AB (Since Tangent at any point of circle is perpendicular to the radius through point of contact)

Now AB is a chord of the circle C₁ and OP⟂AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,

Hence AP = BP