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

(N/A) We are given two concentric circles $C_1$ and $C_2$ with centre $O$ and a chord $AB$ of the larger circle $C_1$ which touches the smaller circle $C_2$ at the point $P$. We need to prove that $AP = BP$.
Let us join $OP$. Then,$AB$ is a tangent to $C_2$ at $P$ and $OP$ is its radius.
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,we have:
$OP \perp AB$
Now,$AB$ is a chord of the circle $C_1$ and $OP \perp AB$. We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
Therefore,$OP$ is the bisector of the chord $AB$,which implies:
$AP = BP$