# Equal Circles, Medial Triangle and Orthocenter

The applet below illustrates the following problem [Altshiller-Court, Theorem 181]:

Any three equal circles having for centers the vertices of a given triangle cut the respective sides of the medial triangle in six points equidistant from the orthocenter of the given triangle.

What if applet does not run? |

### References

- N. Altshiller-Court,
*College Geometry*, Dover, 1980

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Copyright © 1996-2018 Alexander Bogomolny### Solution

Any three equal circles having for centers the vertices of a given triangle cut the respective sides of the medial triangle in six points equidistant from the orthocenter of the given triangle.

**Note**: The problem is equivalent to saying that the six points in question are concyclic; thus it claims the existence of a circle. One can easily surmise that the circle exists even if only four of the six points are available.

Let ΔABC be given, with M_{a}, M_{b}, M_{c} the midpoints and the orthocenter H.

What if applet does not run? |

Assume circle centered at A meets M_{b}M_{c} in U and U'. Let B be the intersection of the altitude from A and M_{b}M_{c}. The two right triangles APU and APU' are equal. Applying the Pythagorean identity gives

PU² = PU'² = AU² - PU² = HU² - HP².

Hence,

AU² - HU² | = AP² - HP² | |

= (AP + HP)(AP - HP) | ||

= AH (DP - HP) | ||

= AH·DH, |

implying

HU² = AU² - AH·DH.

Now, by assumption, AU = BV = CW. So it follows from a property of the orthocenter that *HU = HV = HW.*

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Copyright © 1996-2018 Alexander Bogomolny