Question:

Let $C_1, C_2$ be two concentric circles with common center $O$. $AB$ is a chord of $C_1$ touching $C_2$ at $P$. [0.5 Mark]
$OP$ is the radius of $C_2$. Since $AB$ is tangent to $C_2$ at $P$, $OP \perp AB$. [0.5 Mark]
Now $AB$ is a chord of circle $C_1$ and $OP \perp AB$ from center $O$. [0.5 Mark]
The perpendicular from the center of a circle to a chord bisects the chord. Hence $AP = PB$. Proved! [0.5 Mark]

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🎯 Official CBSE Marking Scheme:
Stating $OP \perp AB$ (tangent property): 1.0 Mark
Applying chord bisector theorem to conclude $AP = PB$: 1.0 Mark