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.

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TopNewestNotice \(\angle AXC = 90\) 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 · 2 years, 3 months ago

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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 · 2 years, 3 months ago

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– Daniel Liu · 2 years, 3 months ago

There may be a more elegant solution.Log in to reply

– Abhishek Sinha · 2 years, 3 months ago

Indeed there is ! Note that the \(\angle APB\) is a right angle. Hence \(PD=\frac{1}{2}BC\).Log in to reply