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Perpendicular (Posted on 2012-12-26) Difficulty: 3 of 5
Let ABC be a triangle and D the foot of the altitude from A. Let E and F lie on a line passing through D such that AE is perpendicular to BE, AF is perpendicular to CF, and E and F are different from D. Let M and N be the midpoints of the segments BC and EF, respectively. Prove that AN is perpendicular to NM.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.3333 (3 votes)

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Solution Analytic Solution | Comment 3 of 5 |

Let a, b, c be real numbers greater than or equal
to zero. Apply a coordinate system to the triangle
such that A = (0,2a), B = (-2b,0), C = (2c,0),
D = (0,0), and M = (c-b,0).
For AEB to be a right angle, point E = (x_e,y_e)
must lie on the circle with diameter AB,
   (x + b)^2 + (y - a)^2 = a^2 + b^2
                   or
         x^2 + y^2 = 2(ay - bx)              (1)
For AFC to be a right angle, point F = (x_f,y_f)
must lie on the circle with diameter AC,
   (x - c)^2 + (y - a)^2 = a^2 + c^2
                   or
         x^2 + y^2 = 2(ay + cx)              (2)
For ANM to be a right angle, point N = (x_n,y_n)
must lie on the circle with diameter AM,
   (x-[c-b]/2)^2 + (y-a)^2 = a^2 + ([c-b]/2)^2
                   or
         x^2 + y^2 = 2ay + (c-b)x            (3)
For the line through D we have 
    y = mx,                                  (4) 
       where
     
    -c/a < m < b/a
For E to be on the line through D it must
satisfy both (1) and (4),
    x_e = 2(am - b)/(1 + m^2)                (5)
For F to be on the line through D it must
satisfy both (2) and (4),
    x_f = 2(am + c)/(1 + m^2)                (6)
For N to be on the line through D it must
satisfy both (3) and (4),
    x_n = (2am + c - b)/(1 + m^2)            (7)
For N to be the midpoint of line segment EF,
it must satisfy the following:
    x_n = (x_e + x_f)/2                      (8)
which it does.
QED


  Posted by Bractals on 2012-12-27 02:40:08
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