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

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.

ÐXPY = ÐXQY = 90^°.

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.

ÐYRX = ÐYSX = 90^°.

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 C1 and C2 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(ÐAXM) = AM R = 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(ÐCYN) = CN r = R XY               (2)

From equations (1) and (2) we have

AM = CN = rR XY

Hence AB = CD.