Let DEF be the orthic triangle of triangle ABC and L,M the centers of the circles passing through D and being tangent at F to the circle (AEF), respectively passing through D and being tangent at E to the circle (AEF). Prove that the circles L, M have BC as radical axis.
Proof:
We need to show that LM_|_BC. Let H be the orthocenter of tr. ABC abd K midpoint of AH. Easy angle calculation show that triangles DFL and DEM are similar, with base angles equal to 90-|B-C|, and one of them only has common points with interior of tr. DEF, consequently, constructing the isosceles triangle FEN, with N and D on either side of EF and similar to DFL and DEM, we know that LFNM is a parallelogram, but easy angle calculation shows FN_|_BC, thus we are done.
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