Sue Liu of Madras College, St Andrew's also sent a very good solution to this problem. Well done Sue!

Conjecture:

AB = CD

Proof We start with Sue's proof that $XY$ is the axis of symmetry of the whole shape.

$XP = XQ$ because they are two tangents from one external point.

$∠ XPY = ∠ XQY = 90^{\circ}$ .

$YP = YQ = r$ so $PYQX$ is a kite and $XY$ is the axis of symmetry of $PYQX$ .

Similarly

$YR = YS$ because they are two tangents from one external point.

$∠ YRX = ∠ YSX = 90^{\circ}$ .

$XR = XS = R$ so $RXSY$ is a kite and $XY$ is the axis of symmetry of $RXSY$ .

Therefore $XY$ is the axis of symmetry of the whole shape.

Sue then goes on to prove in detail that $ABDC$ is a rectangle. Her proof is an excellent piece of work though a little longer than the proof below. The following proof uses sines but it could equally well be written entirely in terms of similar triangles.

The radii of the two circles $C 1$ and $C 2$ are given by:

$XR = XA = R$

$YP = YC = r$ .

Let $M$ and $N$ be the midpoints of the chords $AB$ and $CD$ . Note that $M$ and $N$ are on the line $XY$ joining the centres of the circles and the angles $AMX$ and $CNY$ are right angles.

As $XP$ is a tangent to the circle with centre $Y$ , angle $XPY$ is a right angle.

From the right angled triangles $AXM$ and $YXP$

$\sin (∠ AXM) = \frac{AM}{R} = \frac{r}{XY} (1)$

Similarly $YR$ is a tangent to the circle with centre $X$ and angle $YRX$ is a right angle.

From the right angled triangles $CYN$ and $XYR$

$\sin (∠ CYN) = \frac{CN}{r} = \frac{R}{XY} (2)$

From equations (1) and (2) we have

$AM = CN = \frac{rR}{XY}$

Hence $AB = CD$ .