Prove medians of a triangle can make a triangle.

It means:

If medians are: ma,mb,mcm_a,m_b,m_c, then we have ma+mb>mcm_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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Note that a vector proof (which may be too “algebraic”) is immediate: With DD, EE, FF the midpoints opposite AA, BB, CC, we have →AD=D−A=12(−2A+2B+2C)→BE=E−B=12(−2A−2B+2C)→CF=F−C=12(−2A+2B−2C)\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 →AD+→BE+→CF=0◻\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF} = 0 \qquad\square

– Blue

2 days ago

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2 Answers

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With z∈Cz \in \mathbb{C} being the complex number associated with point ZZ in the complex plane 2⋅ma=|b+c−2a|2 \cdot m_a=|b+c-2a| and similar for mb,mcm_b,m_c, so the inequality to prove becomes: |b+c−2a|+|c+a−2b|≥|a+b−2c||b+c-2a| + |c+a-2b| \ge |a+b-2c|

which follows by the triangle inequality |z1|+|z2|≥|z1+z2||z_1| + |z_2| \ge |z_1+z_2| for z1=b+c−2a\;z_1=b+c-2a,z2=c+a−2b\;z_2=c+a-2b,z1+z2=2c−a−b\;z_1+z_2=2c-a-b.