If you draw a line from the corner of this triangle to the middle of the opposite side, why do the lines intersect at the middle? If you draw a line from the corner of this triangle to the middle of the opposite side, why do the lines intersect at the middle?
This is the triangle (excuse my crude finger painting). I've drawn this for an equilateral triangle, but it holds for other triangles too it seems. This is from Paul Lockhart's book "Measurement", and I thought to ask this here to see if there were any other interesting and intuitive explanations as to why this is.

 A: Consider a triangle ABC. Let D be the midpoint of $\overline{AB}$, E be the midpoint of $\overline{BC}$, F be the midpoint of $\overline{AC}$, and O be the .
By definition, $AD=DB$, $AF=FC$, $BE=EC$ \,. Thus $[ADO]=[BDO]$, $[AFO]=[CFO]$, $[BEO]=[CEO]$, and $[ABE]=[ACE]$ \, where $[ABC]$ represents the area of triangle \triangle $ABC$ ; these hold because in each case the two triangles have bases of equal length and share a common altitude from the (extended) base, and a triangle's area equals one-half its base times its height.
We have:
$$[ABO]=[ABE]-[BEO]$$
$$[ACO]=[ACE]-[CEO]$$
Thus, $[ABO]=[ACO]$ and $[ADO]=[DBO]$, $[ADO]=\frac{1}{2}[ABO]$
Since $[AFO]=[FCO]$, $[AFO]= \frac{1}{2}ACO=\frac{1}{2}[ABO]=[ADO]$, therefore, $[AFO]=[FCO]=[DBO]=[ADO]$. Using the same method, you can show that $[AFO]=[FCO]=[DBO]=[ADO]=[BEO]=[CEO]$.
Source
A: I find this argument intuitive, although it is too elaborate to make a good choice for a formal proof. It didn't turn up in a Google search, nor was it in either of the two books I looked at, so it may be worth posting.
Given a non-degenerate triangle $\Delta = ABC,$ let $\Delta' = A'B'C'$ be its medial triangle. Let this process of forming medial triangles be iterated, thus: $\Delta, \Delta', \Delta'', \ldots, \Delta^{(n)}, \ldots.$ Two iterations are shown here:

By similar triangles, $A''$ is the point where $AA'$ intersects $B'C'.$ Similarly for $B''$ and $C''.$ The important part of this is that $A'A''$ is contained in $AA',$ $B'B''$ in $BB',$ and $C'C''$ in $CC'.$
Iteration thus gives rise to an infinite nested sequence of segments $AA', A'A'', A''A''', \ldots, A^{(n)}A^{(n+1)}, \ldots$ in which each segment is half as long as its predecessor, therefore all the segments intersect in a single point, $G.$
The triangles $\Delta, \Delta', \Delta'', \ldots, \Delta^{(n)}, \ldots,$ considered as including their interior points, also form an infinite nested sequence of closed sets of points, each of which has half the diameter of its predecessor. Therefore, all the triangles intersect in a single point; and this point must be $G,$ because it is contained in all the triangles.
By the same argument, $G$ is also the point of intersection of $BB', B'B'', B''B''', \ldots, B^{(n)}B^{(n+1)}, \ldots,$ and at the same time, it is the point of intersection of $CC', C'C'', C''C''', \ldots, C^{(n)}C^{(n+1)}, \ldots.$
Therefore, the segments $AA', BB', CC'$ meet internally in the point $G.$
