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Prove medians of a triangle can make a triangle. It means: If medians are: $m_a,m_b,m_c$, then we have $m_a + m_b > m_c$.

I know we can prove it using the length of medians (Apollonius theorem) but I want a geometric prove, not using pure algebra and square roots and similar things.

Thanks and sorry for my English.

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    $\begingroup$ Note that a vector proof (which may be too "algebraic") is immediate: With $D$, $E$, $F$ the midpoints opposite $A$, $B$, $C$, we have $$\begin{align}\overrightarrow{AD} &= D-A = \frac{1}{2}(-2A+\phantom{2}B+\phantom{2}C) \\ \overrightarrow{BE} &= E - B = \frac{1}{2}(\phantom{-2}A-2B+\phantom{2}C) \\ \overrightarrow{CF} &= F - C = \frac{1}{2}(\phantom{-2}A +\phantom{2}B - 2 C)\end{align}$$ so that $$\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = 0 \qquad\square$$ $\endgroup$ – Blue Oct 21 '16 at 21:44
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With $z \in \mathbb{C}$ being the complex number associated with point $Z$ in the complex plane $2 \cdot m_a=|b+c-2a|$ and similar for $m_b,m_c$, so the inequality to prove becomes: $$|b+c-2a| + |c+a-2b| \ge |a+b-2c|$$ which follows by the triangle inequality $|z_1| + |z_2| \ge |z_1+z_2|$ for $\;z_1=b+c-2a$,$\;z_2=c+a-2b$,$\;z_1+z_2=2c-a-b$.

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