Here's a picture because I can't really explain the problem. Tried to prove it with using the fact that the centroid divides a median in a ratio $2:1$ and similar triangles but I got stuck.
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Hint: Let $P$ be the midpoint of $CB$. Drop a perpendicular from $P$ to that "line under" – Lozenges Nov 25 '17 at 14:48
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@Lozenges I tried it but I don't see what I should do next. The only thing I got is that the line M to the side AB is equal to 2/3 P to the side AB and that C to the side AB is 2 times the line P to the side AB. Not sure if that is sufficient – pavle Nov 25 '17 at 15:03
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Ayy Lmao, have you managed to prove that the length of the perpendicular is $(d_2+d_3)/2$? – Nov 25 '17 at 15:15
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@user8734617 No, how would I prove that? – pavle Nov 25 '17 at 15:18
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Median of a trapezoid? – Nov 25 '17 at 15:19
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Oh, you're right, I don't know how I didn't notice that. – pavle Nov 25 '17 at 15:21
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I still don't know what I could use that for – pavle Nov 25 '17 at 15:29
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Let $x_1,x_2,x_3$ be the x coordinates of points A, B and C respectively. So point A is $ (x_1,d_1)$, point B is $(x_2,d_2)$ and point C is $(x_3,d_3)$. Now M is the centroid of triangle. Therefore y-coordinate of M is $$(\frac{d_1+d_2+d_3}{3})$$ hence proved that $$MD=(\frac{d_1+d_2+d_3}{3})$$
Rohan Shinde
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Yes, but can you prove it without coordinate geometry, I know that there is a formula for the centroid of a triangle, but I would really like to know how to prove it using "normal" euclidean geometry. – pavle Nov 25 '17 at 15:16
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OK sorry I'll just try coz I don't have much idea of Euclidean Geometry – Rohan Shinde Nov 25 '17 at 15:17
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