+0  
 
+1
157
1
avatar+1442 

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.

 

Thanks!

AnonymousConfusedGuy  Jun 18, 2018
 #1
avatar
0

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.

 

Tiggsy.

Guest Jun 19, 2018

43 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.