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.
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
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Posted by Bractals
on 2012-12-27 02:40:08 |
Analytic Solution
