Posted by **Amy** on Sunday, November 30, 2008 at 1:39pm.

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**, Sunday, November 30, 2008 at 2:01pm
You are dealing with the intersection of the three altitudes, which is the orthocentre.

The third altitude, from B to the x-axis 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 ÷ 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.

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