### Orthopole: What is it?

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Copyright © 1996-2018 Alexander BogomolnyGiven ΔABC and a straight line m. First drop perpendiculars Aa, Bb and Cc from the vertices of ΔABC onto m. From the three points thus obtained drop perpendiculars on the "opposite" sides of the triangle: from a onto BC, from b onto AC, and from c onto AB. The latter three lines intersect at a point, known as the *orthopole* of ΔABC and m.

Let P be the point of intersection of the perpendiculars from a to BC and from b to AC. In triangles ACD and bPa the sides are pairwise orthogonal: AC ⊥ Pb, AD ⊥ ab, and CD ⊥ Pa. From here

(1) | AD/CD = ab/Pa |

Similarly, let Q be the point of intersection of the perpendicular from a to BC and from c to AB. In triangles ABD and cQa the sides are pairwise orthogonal: AB ⊥ Qc, BD ⊥ Qa, and AD ⊥ ac. From here

(2) | BD/AD = Qa/ac |

Additionally, since Aa||Bb||Cc,

(3) | CD/BD = ac/ab |

Multiplying (1-3) we get

1 = AD/CD·BD/AD·CD/BD = ab/Pa·Qa/ac·ac/ab = Qa/Pa, |

or Pa = Qa. Since both P and Q lie on the same perpendicular from a to BC, P = Q, which establishes the concurrency of the three perpendiculars.

(Darij Grinberg came up with a short proof based on a theorem of Carnot.)

### References

- R. Honsberger,
*Episodes in Nineteenth and Twentieth Century Euclidean Geometry*, MAA, 1995.

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