$O$ is the centre of the large circle
$AB$ is a chord of the large circle
$OB$ is a diameter of the small circle.
Both circles touch at $B$
The small circle cuts the chord $AB$ at $X$
prove that $AX = XB$
Attempt
I made $OB$ and $AO$ into lines which created an isosceles triangle (because $O$ is the midpoint). I think the next step is to make a line to bisect it, but how would I know where to draw the line?
Thank heaps.
We have: $\angle OXB = 90^{\circ}$ because $OB$ is the diameter of the smaller circle, thus $\triangle OAX \cong \triangle OBX$ using hyp-leg criteria in right triangles. It follows that $XA \cong XB$ as CPCTC.
$OA=OB$ and also $OX$ is the perpendicular to $AB$ therefore it is the perpendicular bisector of $OAB$, thus $AX=BX$
A homothety centered at $B$ taking the small circle to the large circle has scale factor $2$, since $O$ goes to the antipode of $B$ in the large circle. This implies the result.
