Concurrency of Circles and Sides

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While playing around on Geogebra, I found a curious theorem:

For all triangles ABC\triangle ABC, draw the circles with diameter as each of its sides. Call the circle passing through points AA and BB circle OCO_C, and ditto for the other two circles. This theorem states that circles OAO_A, OBO_B, and line ABAB are concurrent, and ditto for the other two cases.

Your challenge: prove this theorem! Also, what is the significance of the concurrency points X,Y,ZX,Y,Z? Is there a simpler way to define these points?

If this is actually a real theorem, please point me to the name of it.

Note by Daniel Liu
7 years ago

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Notice AXC=90\angle AXC = 90 due to ABAB being the diameter. Similarly CXB=90\angle CXB = 90, so AXB=180\angle AXB = 180 meaning XX is on ABAB. Similarly Y,ZY, Z are on CA,BCCA, BC.

The altitudes result in AX,BY,CZAX, BY, CZ being the altitudes of ABC\triangle ABC, concurring at the orthocenter and forming the orthic triangle of ABCABC (as a result, A,B,CA, B, Care the excenters of XYZ\triangle XYZ)

Akshaj Kadaveru - 7 years ago

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Hint : Restate the theorem a bit, Let the circle OAO_A and the line-segment ACAC meet at point PP. Join point PP to the midpoint of line-segment ABAB (call it DD) . We need to prove that DP=AD=c2DP=AD=\frac{c}{2}. The trigonometric formula b=acosC+ccosAb=a\cos C + c\cos A will be useful.

Abhishek Sinha - 7 years ago

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There may be a more elegant solution.

Daniel Liu - 7 years ago

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Indeed there is ! Note that the APB\angle APB is a right angle. Hence PD=12BCPD=\frac{1}{2}BC.

Abhishek Sinha - 7 years ago

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