O is the centre of the small circle. CA and CB are tangents to this circle. The large circle is drawn through A, B and C. CO is produced to E and intersects the small circle at D.
i. Show that O lies on the large circle
ii. Show that CD = (CB x CA) / (CD + 2DO)
Let Q be the point on CO at which the larger circle intersects CO. We want to show that this O.
Since CQ is a diameter of the larger circle, angle CAQ is a right angle (standard theorem in geometry). So as CA is a tangent of the smaller circle, and is perpendicular to AQ, it follows that AQ must be a radius of the smaller circle and so AQ = AO.
So Q = O.
That's the first bit done.
I think the picture should look like the attachment. Follow Matt Westwood's advice for part (i). For (ii), notice that CB = CA and CE = CD + 2DO (assuming that E is the point where the line CDO crosses the small circle again). Then the result CD = (CB x CA) / (CD + 2DO) is equivalent to . This is a standard result which you can prove for example by using the fact that the triangles ADC and EAC are similar.
Multiply out the fraction. Then the equation CD = (CB x CA) / (CD + 2DO) becomes CD × (CD + 2DO) = CB × CA. But you can see from the picture that CE = CD + DO + OE, where DO and OE are both equal to the radius of the small circle. So CE = CD + 2DO. Also, CB = CA (because the two tangents from a point to a circle are always equal).