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$O$ 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 $AX = XB$

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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.

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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.

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$OA=OB$ and also $OX$ is the perpendicular to $AB$ therefore it is the perpendicular bisector of $OAB$, thus $AX=BX$

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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.

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