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.