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what is the length of side AB in triangle ABC with AB=AC, BC=8, and median CD=9

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  1. I will assume that D is the midpoint of AB
    draw the median from A to BC to meet BC at E
    let the intersection of these two medians be F
    Since you have an isosceles triangle that median will meet BC at right angles.
    Also the medians intersect each other in the ratio of 2:1, the longer side towards the vertex.
    Then FC = 6
    by Pythagoras
    FE^2 + 16 = 36
    FE = √20
    then AE = 3√20

    finally AC^2 = AE^2 + EC^2
    = 180 + 16
    = 196
    AC = √196 = 14
    But AB = AC
    so AB = 14

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