If triangles {ABC,OO'A} are similar, then CO'DA is cyclic, D being the intersection point of {OO',BC}.

But O'CA is a right angle, hence O'DA is a right angle too.

Inversely assuming the above configuration, CO'DA is again cyclic. This time because at C, D are right angles. This implies that the angles at O', C are equal.

But ABOD is then also cylic, hence angle BAO is equal to BDO which is equal to CAO'. This implies that the angles at A of ABO' and AOO' are equal.

The circle on diameter H

The equality of the two segments is equivalent to the equality of powers of {C,D} with respect to the circles {k

This implies that the projections {B',E'} of {B,E} on {AD,AC} respectively, together with {D,C} build a cyclic quadrilateral. Hence triangles CAB' and DAE' are similar and by the cyclic quadrangles {DAE'E, BB'AC} also the right-angled triangles AEE' and ABB' are similar. This implies that AE/AB = r

This follows trivially by drawing lines {AE, AB} and calculating their ratio in terms of the radii of the two circles.

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