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.
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 :)
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1 answer