Question

Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle, is bisected at the point of contact.

Answer 1

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.

Therefore, $OP\perp AB$.

But, the perpendicular drawn from the center of a circle to a chord bisects the chord.

Therefore, OP bisects AB.

Hence, AP = BP.