# Why is the Centroid of a Triangle Shared with Its Midpoint Triangle

The question is as follows:

Let $A'$, $B'$ and $C'$ be the midpoint triangle of triangle $ABC$. In other words, $A'$, $B'$ and $C'$ are the midpoints of segments $BC$, $CA$ and $AB$, respectively. Show that triangles $A'B'C'$ and $ABC$ have the same centroid.

My reasoning was through the means of the Midline Theorem. After drawing the diagram, we can see, for example, $C'A'$ is a midline of $\triangle ABC$, which means that $C'A'$ is parallel to the base $AC$ and half the length of it as well. Therefore, the median $B'B$ would pass through the midpoint of $C'A'$, including the midpoint of $AC$.

Is my reasoning valid or, at least, is it on the right track? Any advice on solving this problem will be greatly appreciated.

• You're missing a nuance. The midline being parallel to (and half the length of) the base doesn't give you the median property immediately. However, it does give you some similar triangles to play with, and a straightforward proportionality argument can get you where you need to go. (Another approach: Consider two of the midlines and the quadrilateral they make the corresponding halves of the triangle's sides.)
– Blue
Commented Sep 17, 2017 at 3:39
• "′C′A′ is parallel to the base AC and half the length of it" does not immediately imply "the median B′B would pass through the midpoint of C′A′, including the midpoint of AC". More work need to be shown to convince others that the median of A'B'C' is part of that of ABC.
– Mick
Commented Sep 17, 2017 at 3:41
• @Blue I am seeing the formation of parallelograms in the process. $C'B'A'C$ forms a parallelogram with $C'B' \parallel A'C$ and $C'A' \parallel B'C$. Am I looking at it through the perception that you're telling for me to look at it as? Commented Sep 17, 2017 at 3:45
• @geo_freak: Yes, parallelograms are what I was suggesting. In the one you mentioned, think about what you want to prove: The median from $C$ in $\triangle ABC$ is also a median of $\triangle A^\prime B^\prime C$; that is, that $CC^\prime$ and $A^\prime B^\prime$ have a certain special relation. (BTW: I didn't mean to distract you with the parallelogram stuff. It's actually worthwhile for you to try to fill the gap in your original argument.)
– Blue
Commented Sep 17, 2017 at 4:09
• @Blue I was trying to fill the gaps with my original argument, however, I still am unsure about what to do with the similar triangles. How can I use the similar triangles to prove my point? Any hint will be appreciated. Commented Sep 17, 2017 at 4:55

Let $A(a)$, $B(b)$ and $C(c)$ in the Gauss's plane.

Thus, $M\left(\frac{a+b+c}{3}\right)$ is a centroid of $\Delta ABC$ or $$M\left(\frac{\frac{b+c}{2}+\frac{a+c}{2}+\frac{a+b}{2}}{3}\right),$$ which is a centroid of $\Delta A'B'C'$.

Done!

• What is "Gauss's plain" (I suppose you mean "plane")? Commented Sep 17, 2017 at 6:24
• @Wojowu Thank you! I fixed. Commented Sep 17, 2017 at 6:26
• So what is Gauss's plane? Commented Sep 17, 2017 at 6:27
• @Wojowu See here: en.wikipedia.org/wiki/Complex_plane Commented Sep 17, 2017 at 6:28
• Thanks, I never knew the complex plane is called so (the Wikipedia article doesn't seem to indicate that, unless I'm missing it). Commented Sep 17, 2017 at 6:30

Observe a homothety with center at the centroid $T$ of the triangle $ABC$ and $k=-{1\over 2}$. This homothety takes $A$ to $A'$ and so on. Thus it takes centroid $T$ to the centroid $T'$ of $A'B'C'$ (beacuse a homothety is a similarity transformation). But the center of homothethy is fixed point for a homothety, so $T'=T$.