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25 June, 13:08

Prove that in a triangle, the length of every median is smaller than the average of the lengths of the two sides meeting at the median's vertex.

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  1. 25 June, 14:15
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    Given: Consider a triangle ABC in which AD is median drawn from vertex A.

    To prove: AB + AC > AD

    Proof: In Δ A B D

    AB + B D> AD ⇒[In a triangle sum of lengths of two sides is greater than the third side] ... (1)

    In Δ A CD

    AC + DC > AD [In a triangle sum of lengths of two sides is greater than the third side] ... (2)

    Adding (1) and (2)

    AB + AC + B D + D C > B D + DC

    A B + A C + B C > 2 B D ... (3)

    Also, Considering Δ AB C

    AB + B C > B C ⇒[In a triangle sum of lengths of two sides is greater than the third side]

    ⇒ AB + B C - B C >0 ... (4)

    Adding (3) and (4)

    A B + B C+B C + AB + A C - B C > 2 A D

    ⇒2 AB + 2 A C> 2 A D

    Dividing both side of inequality by 2, we get

    A B + A C> A D

    Hence proved.
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