In triangle ABC, AB = AC, D is the midpoint of line BC, E is the foot of the perpendicular from D to line AC, and F is the midpoint of line DE. Prove that line AF is perpendicular to line BE.



AnonymousConfusedGuy  Jun 18, 2018

Easiest solution, I think, is via co-ordinate geometry.


Align the triangle so that A is at the origin and AC lies along the positive x-axis. (DE will then be perpendicular to the x-axis.)


Let the co-ordinates of B be (4a, 4b) and the co-ordinates of C be (4c, 0).

(The 4's are there so as to avoid fractions later in the calculation.)

Since AB = AC in length, (4a)^2 + (4b)^2 = (4c)^2, so a^2 + b^2 = c^2,

in which case b^2 = -(a^2 - c^2), (needed later).

D is the mid-point of BC so its co-ordinates will be (2a + 2c, 2b), in which case the co-ordinates of E will be (2a + 2c, 0),

and since F is the mid-point of DE, its co-ordinates will be (2a + 2c, b).


Now onto the slopes of the two lines.

The slope of AF will be b/(2a + 2c) and the slope of BE (4b - 0)/(4a - (2a + 2c)) = 4b/(2a - 2c).

The product of these two will be

\(\displaystyle \frac{b}{2(a + c)}.\frac{4b}{2(a-c)}=\frac{b^{2}}{a^{2}-c^{2}}=-1, \)

so the two lines are at right-angles. QED.


The problem can also be solved using vectors and also by a combination of Pythagoras and similar triangles.



Guest Jun 19, 2018

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