Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose you have a triangle $ABC$ and a point $P$ which lies on the interior of the triangle.

Form three subtriangles by connecting $P$ to the vertices of $ABC$. Show that if the three subtriangles have equal area, then $P$ is the centroid. Can someone give me a hint?

share|cite|improve this question
What definition do you have for the centroid? What facts/theorems do you know about the centroid? Easier to help you when we know what you know. – Gerry Myerson Mar 26 '12 at 6:15
Centroid is the intersection of the 3 medians. Only 2 theorems I know is that that centroid lies 2/3 along median, and the medial triangle has same centroid. – StuartHa Mar 26 '12 at 6:18
up vote 1 down vote accepted

Because $\triangle PAB$ has area $1/3$ of the area of $\triangle ABC$, the height of $\triangle PAB$, with respect to the base $AB$, is $1/3$ of the height of $\triangle ABC$ with respect to $AB$. Thus $P$ lies on the line $l$ parallel to $AB$, such that the distance between $l$ and $AB$ is $1/3$ of the distance from $C$ to $AB$.

For the same reason, $P$ lies on the line $m$ parallel to $BC$, such that the distance between $m$ and $BC$ is $1/3$ of the distance from $A$ to $BC$.

But because the centroid of $\triangle ABC$ divides each of the medians in the ratio $2:1$, it follows that the centroid also lies on both $l$ and $m$. Since there is only one point which is on both $l$ and $m$, the centroid must be $P$.

share|cite|improve this answer

Each of the three subtriangles has an area a third of the original triangle and a base which is an edge of the original triangle. So their perpendicular heights must be a third of the originals (the originals being the medians).

This means $P$ must lie on each of the three lines through the centroid parallel to the edges (perpendicular to the medians). The only point which lies on all three of these lines (or indeed any two of them) is the centroid.

As an aside, the lines through the centroid parallel to the edges do not divide the triangle in to two equal areas, but instead into a triangle and trapezium with areas $\frac49 : \frac59$.

share|cite|improve this answer

You write that you know that the centroid is the intersection of the three medians. Since the medians go through the vertices and the lines connecting $P$ to the vertices go through $P$ and the vertices, if $P$ is to be the centroid, these lines have to be the medians, and conversely if you can show that they're the medians it will follow that $P$ is the centroid.

share|cite|improve this answer
Isn't that just restating the definition I gave? – StuartHa Mar 26 '12 at 6:37
@user19192: Sorry, sometimes it's hard to tell what's obvious for people and what isn't and where the border between a hint and a solution lies -- I guess I was being a bit overcautious because you said you were only looking for a hint :-) – joriki Mar 26 '12 at 8:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.