Let $p$ be a line that pass through the centroid of a triangle $ABC$. Unless the line pass through one vertex, then $2$ verices are one side of the line, while the third one is on the other side. Without loss of generality, let vertices $A$ and $C$ be one side, while the vertex $B$ be on the other side of the line. Let $x,y,z$ be the distances from the vertices $A,B,C$ to the line $p$. Prove that $x + z = y$

I've tried something, but I didn't make any progress. If we multiply both sides of the equality by $\frac{\overline{XY}}{2}$, then we need to prove:

$$P_{BXY} = P_{AXY} + P_{CXY} \iff \frac{XB \cdot BY \cdot \sin \angle XBY}{2} = \frac{XA \cdot AY \cdot \sin \angle XAY}{2} + \frac{XC \cdot CY \cdot \sin \angle XCY}{2} \iff XB \cdot BY = XA \cdot AY + XC \cdot CY$$

But this is the most that I manage to get.

  • $\begingroup$ You haven't used the fact that $p$ passes through $M$ yet, although you will need it somewhere. What are some nice properties of the centroid that may be useful here? $\endgroup$
    – Ragnar
    Jan 30, 2014 at 19:00

1 Answer 1


Let $F$ the the midpoint of line segment $AC$. We know that $F$ lies on $BM$, because $M$ is the centroid. Let $G$ be the point of intersection of $p$ and the perpendicular to $p$ through $F$. It's not hard to see that $$ |FG|=\frac{|AI|+|CJ|}2 $$ although I can't see a nice simple proof of it right now. (Found one, see below) Because $M$ is the centroid, we know that $|BM|=2|MF|$. Now, we have that $\triangle BMH\sim\triangle FMG$, because of equal angles en parallel lines. It follows that $$ \frac{|FG|}{|BH|}=\frac{|FM|}{|BM|}=\frac 12 $$ Combining this with the earlier found equation, we get that $|AI|+|CJ|=|BH|$ indeed.

To show the first equality, you can look at the triangles with sides $AF$ and $CF$, parts of the line through $F$ parallel to $p$ and the corresponding parts of the perpendicular through $A$ and $C$ on $p$. The two triangles you obtain are congruent, so the lengths of the excess at $A$ and the shortage at $C$ are the same.

  • 1
    $\begingroup$ Quite nice proof. I tried to use the fact the M is te centroid and splits $BF$ in ratio $2:1$, but I couldn't make any use of it. $\endgroup$
    – Stefan4024
    Jan 30, 2014 at 19:12

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