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 20101025 15:27:55 