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# Concurrency of Circles and Sides

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

For all triangles $$\triangle ABC$$, draw the circles with diameter as each of its sides. Call the circle passing through points $$A$$ and $$B$$ circle $$O_C$$, and ditto for the other two circles. This theorem states that circles $$O_A$$, $$O_B$$, and line $$AB$$ are concurrent, and ditto for the other two cases.

Your challenge: prove this theorem! Also, what is the significance of the concurrency points $$X,Y,Z$$? 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
2 years, 8 months ago

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Notice $$\angle AXC = 90$$ due to $$AB$$ being the diameter. Similarly $$\angle CXB = 90$$, so $$\angle AXB = 180$$ meaning $$X$$ is on $$AB$$. Similarly $$Y, Z$$ are on $$CA, BC$$.

The altitudes result in $$AX, BY, CZ$$ being the altitudes of $$\triangle ABC$$, concurring at the orthocenter and forming the orthic triangle of $$ABC$$ (as a result, $$A, B, C$$are the excenters of $$\triangle XYZ$$) · 2 years, 8 months ago

Hint : Restate the theorem a bit, Let the circle $$O_A$$ and the line-segment $$AC$$ meet at point $$P$$. Join point $$P$$ to the midpoint of line-segment $$AB$$ (call it $$D$$) . We need to prove that $$DP=AD=\frac{c}{2}$$. The trigonometric formula $$b=a\cos C + c\cos A$$ will be useful. · 2 years, 8 months ago

There may be a more elegant solution. · 2 years, 8 months ago

Indeed there is ! Note that the $$\angle APB$$ is a right angle. Hence $$PD=\frac{1}{2}BC$$. · 2 years, 8 months ago