The proof follows at once from the fact that quadrangles DFEB and DCHB are cyclic. Hence angle(CHD) = angle(CBD) = angle(HEF). Showing that EF and CH are parallel.

This has an interesting consequence studied in SimsonProperty2.html .

Draw from C a line CH in the given direction and determine point H. Then from H draw the orthogonal to the oposite side (AB) intersecting the circle at a second point D. This is the desired point.

The proof follows from the equality of triangles DEF and GPA which is trivial and the fact that E is symmetric to H with respect to BC.

This argument is used also to prove a property of orthopoles (see Orthopole.html ).

Orthocenter.html

Orthopole.html

Projection_Triangle.html

Simson.html

SimsonDiametral.html

SimsonGeneral.html

SimsonGeneral2.html

SimsonProperty2.html

Simson_3Lines.html

SimsonVariantLocus.html

SteinerLine.html

ThreeDiameters.html

ThreeSimson.html

TrianglesCircumscribingParabolas.html

WallaceSimson.html

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