Two smaller circles of radii a and b are internally tangent to a larger circle, of radius r, at points P and Q respectively. The smaller circles intersect at points S and T.
If P, S, and Q are collinear, prove that r = a+b.
Call the centers of the circles with radius a, b and r, A, B and R respectively.
Triangles PAS, SPQ and PRQ are isosceles (having a circle's center and two ends of a chord as vertices). Therefore angle APS = angle PSA, and similarly for angles BSQ = SQB and RPQ = PQR. But angle APS is the same as (coincident with) angle RPQ, and RQS is the same as PQR so all these triangles are similar.
Since angle PSA = angle SQB, line AS is parallel to line QBR and quadrilateral ASBR is a parallelogram, making segment SB, which is the radius of circle B, equal to segment AR, which together with AP, the radius of circle A, makes up segment RP, the radius of circle R. QED.
Posted by Charlie
on 2009-07-13 17:53:05