Two circles touch internally at a point A and the smaller of the two circles passes through O, the centre of the larger circle. AB is any chord of the larger circle, cutting the smaller circle S. The tangents to the larger circle at A and B meet at a point T.

Prove:

i) AB is bisected at S.

ii) O, S and T are collinear.

2. Hello, xwrathbringerx!

Two circles touch internally at a point $\displaystyle A$
and the smaller of the two circles passes through $\displaystyle O$, the centre of the larger circle.
$\displaystyle AB$ is any chord of the larger circle, cutting the smaller circle $\displaystyle S.$
The tangents to the larger circle at $\displaystyle A$ and $\displaystyle B$ meet at a point $\displaystyle T.$

Prove:

a) $\displaystyle AB$ is bisected at $\displaystyle S.$

b) $\displaystyle O, S\text{ and }T$ are collinear.

Draw $\displaystyle OT.$
Draw radii $\displaystyle OA = OB = r$

$\displaystyle TA = TB$
Tangents to a circle from an extrenal point are equal.

Points $\displaystyle O$ and $\displaystyle T$ are equidistant from points $\displaystyle A$ and $\displaystyle B.$
. . Hence, $\displaystyle OT$ is the perpendicular bisector of $\displaystyle AB.$

$\displaystyle OA$ is a diameter of the small circle.
$\displaystyle \angle OSA$ is inscribed in a semicircle: .$\displaystyle \angle OSA = 90^o$

Hence, $\displaystyle S$ lies on $\displaystyle OT.$

And the two proofs follow . . .

3. Can I simply say OA is a diameter of the small circle?