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Orthopole: What is it?
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(All blue elements - the points, the line and the triangle are draggable.)
Given 
ABC and a straight line m. First drop perpendiculars Aa, Bb and Cc from the vertices of ABC 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 ABC and m.
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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
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