Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle, is bisected at the point of contact.
Given: Two circles with the same centre O and AB is a chord of the larger circle which touches the smaller circle at P.
To prove: AP = BP.
Construction: Join OP.
Proof: AB is a tangent to the smaller circle at the point P and OP is the radius through P.
∴ OP ⊥ AB.
But, the perpendicular drawn from the centre of a circle to a chord bisects the chord.
∴ OP bisects AB. Hence, AP = BP