Let AD be the altitude perpendicular to BC, let BE be the altitude perpendicular to AC and let AD meet BE at H. Finally let C H meet AB at F. We must prove FC is perpendicular to AB.
Join DE. Let α = ∠BC F. Then ECDH is a cyclic quadrilateral (supplementary opposite angles), so ∠DE H = α (angles in the same segment).
Similarly, AEBD is a cyclic quadrilateral and ∠D AB = α.
From 4ABD, ∠ABD = 180◦ −90◦ −α = 90◦ −α.
Finally, in 4CBF, ∠BFC + α + 90◦ − α = 180◦, so ∠BFC = 90◦, as required.
