ABC is an equilateral triangle. M is a point inside triangle ABC and, the respective feet of perpendiculars from M to the sides BC, CA and AB are denoted by D, E and F.

Determine the locus of all such points M for which / FDE is a right angle.

Let /XYZ denote angle XYZ.

Let O be the center of triangle ABC and
N the intersection of the ray AO and the
bisector of /CBO. The locus of points M 
is the circular arc BNC.

PROOF:

Clearly /BCN = /BNC = 15. Thus,
/BMC = /BNC = 150.

   /FDE = /FDM + /MDE
        
        = /FBM + /MCE

        = (60 - /MBC) + (60 - /MCB)

        = 120 - (/MBC + /MCB)

        = 120 - (180 - /BMC)

        = 120 - (180 - 150)

        = 90.

Posted by Bractals on 2010-10-25 15:27:55