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(All blue elements - the points, the line and the triangle are draggable.)

Explanation

Given 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

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

 (3) CD/BD = ac/ab

Multiplying (1-3) we get