The figure above (due to L. Fejer) suggests the solution. Assume a point D on BC fixed and consider all triangles DFE with variable E, F only. Reflect D on the sides to obtain points D', D'' forming the isosceles D'AD'' with angle(D'AD'') = 2*A and legs of length equal to |AD|. Obviously, among these triangle the one with least perimeter is DE'F', its perimeter being D'D''. Now vary also point D and seek the triangle DF'E' with least perimeter. Since the perimeter is |D'D''| = 2*|AD|*sin(A), this becomes minimal when AD coincides with the altitude from A. Then DE'F' takes the position of the orthic triangle and |D'D''| becomes equal to the perimeter of the orthic triangle.

See the file Fagnano.html for another proof relating also to the billiard-ball problem.

The figure below displays the right position for D giving the solution to the problem.

Orthic.html

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