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difficult geometry problem. help is appreciated :)

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Circles $$S$$ and $$T$$ have radii $$1$$ and intersect at $$A$$ and $$B$$. The distance between their centers is $$\sqrt{2}$$. Let $$P$$ be a point on circle $$S$$, and let $$\overline{PA}$$ and $$\overline{PB}$$ intersect circle $$T$$ again at $$C$$ and $$D$$, respectively. Show that $$\overline{CD}$$ is a diameter of circle $$T$$.

Thanks a lot!

Apr 3, 2020

1+0 Answers

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Let L be the center of circle S, and let F be the center of circle T.  Since LA = AF = 1 and LF = sqrt(2), triangle LAF is a 45-45-90 triangle.  This means angle ALF is 45 degrees.  By symmetry, angle BLF is also 45 degrees.

Then angle ALB = angle ALF + angle LBF = 45 + 45 = 90 degrees, so arc AB is 90 degrees.  Since angle APB subtends arc AB, angle APB is 90/2 = 45 degrees.  Then by Poncelet's Theorem, triangle PCD is a 45-45-90 triangle, so angle PCD is 90 degrees.  Then angle ACD is 90 degrees, which implies that CD is a diameter of circle T.

Apr 3, 2020