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.

#Geometry #Triangle #Concurrency #Midpoint #Circles

Easy Math Editor

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Comments

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 $\triangle XYZ$ )

Akshaj Kadaveru - 6 years, 11 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.

Abhishek Sinha - 6 years, 11 months ago

There may be a more elegant solution.

Daniel Liu - 6 years, 11 months ago

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

Abhishek Sinha - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$