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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?

Amy :)

  • Geometry -

    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)

    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

    follow the same steps to show M is the midpoint of AB, for the given equilateral condition.

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