# 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
5 years, 1 month 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 $\triangle XYZ$)

- 5 years, 1 month 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.

- 5 years, 1 month ago

There may be a more elegant solution.

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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