Posted by Amy on .
Do the altitudes of an isosceles triangle go to the midpoints of the opposite sides?
My triangle ABC is isosceles with points A (0,0), B (a,b), and C (2a,0). My diagram shows that the altitude from A will hit side BC at point N and the altitude from C will hit side AB at point M, but I can't find anywhere in the book if M and N are supposed to be midpoints. Help?
Thanks,
Amy :)

Geometry 
Reiny,
You are dealing with the intersection of the three altitudes, which is the orthocentre.
The third altitude, from B to the xaxis will definitely hit the midpoint of AC.
The others will only hit the midpoint of AB and BC if the triangle is equilateral.
for that to happen AB = BC
√(b^2 + a^2) = 2a
b^2 + a^2 = 4a^2
b = √3a, so point B must be (a,√3a)
Proof:
If N is the midpoint then the product of the slopes of BC and AN must be 1
(their slopes must be negative reciprocals of each other)
N is midpoint of BC = (3a/2,√3a/2)
slope of BC = √3a/a = √3
slope of AN = √3a/2 รท 3a/2 = √3/3
product = √3(√3/3) = 1
QED
follow the same steps to show M is the midpoint of AB, for the given equilateral condition.