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Equilateral Right Angle Locus (Posted on 2010-10-25) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Solution | Comment 1 of 2

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
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