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ABC is a triangle with AC=139 and BC=178. Points D and E are the midpoints of BC and AC respectively. Given that AD and BE are perpendicular to each other, what is the length of AB?

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    Since D, E bisect BC and AC respectively, we have
    AE=139/2=69.5
    BD=178/2=89

    Let medians AD and BE meet at K.
    Let KD=x, DE=y
    Then
    BK=2y, AK=2x
    since medians intersect each other at third points.

    Consider right triangle AKE right angled at K:
    (2x)²+y²=69.5² (Pythagoras)
    =>
    4x²+y²=69.5²....(1)

    Consider right triangle BKD,
    x²+4y²=89².....(2)

    4(1)-(2) to eliminate y:
    15x²=4*69.5²-89²
    =>
    x²=760

    Substitute x² in (2):
    4y²=89²-760=7161

    Finally, consider right triangle AKB:
    AB²
    =4x²+4y²
    =3040+7161
    =10201
    =>
    AB=101

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