In the given figure, the chord AB of the larget of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.
Let there is a circle having center O
Let AB is the tangent to the smaller circle and chord to the larger circle.
Let P is the point of contact.
Now, draw a perpendicular OP to AB
Now, since AB is the tangent to the smaller circle,
So, ∠OPA = 90
Now, AB is the chord of the larger circle and OP is perpendicular to AB.
Since the perpendicular drawn from the center of the circle to the chord bisect it.
So, AP = PB
Hence, in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact