The proof is a consequence of the fact that the parallelogram p = EFGH, with vertices at the middles of the sides is a rectangle. Then the feet, as for example L, view a diagonal [FH] of p under a right angle etc.

Notice that these perpendiculars intersect mutually on the diagonals of the initial quadrilateral. This is seen, for example at P, from the fact that angle(PAK) = angle(PIK) = angle(DGH), pursuing the cyclic quadrilaterals AKPI and KGHI. Thus, by parallelity AP is perpendicular to FG and must pass through M. See the file Orthodiagonal.html for a further discussion of the subject.

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