Half, Half, Half

Consider points $D$ & $E$ to be the midpoints of $AB$ & $AC$ respectively, as shown above. Then the $\triangle ADE$ will share one same angle and has 2 sides of half the ratio as the $\triangle ABC$ . In other words, $\triangle ADE$ is similar to $\triangle ABC$ . As a result, the segment $DE$ will be half of $BC$ due to similarity ratio.

Hence, if we join the midpoints of $\triangle ABC$ in similar fashion, all the three sides of $\triangle DEF$ will be half to those of $\triangle ABC$ . Therefore, the ratio among the three sides will be the same in both triangles; both triangles are similar.

According to the Mid point theorem, if we join the mid-points of two sides of the triangle, the resulting line segment is parallel to the third side of the triangle and is equal to $1/2$ of the the third side.

Let us say that the larger triangle is $\Delta ABC$ and the smaller one is $\Delta DEF$ . Thus, applying mid point theorem to all three corresponding sides we arrive at:

$\dfrac{AB}{DF}=\dfrac{BC}{EF}=\dfrac{AC}{ED}=\dfrac 12$

Now using the SSS criterion for similar triangles we have proved that they are similar.

ABC be a triangle. DEF form another triangle such that D, E, F are midpoints of AC, BC, AB resp. Since, Lines, DE, FE, and FD are passing through midpoints, by mid-point theorem DE || AB, FE || AC, FD || BC. In such a case, AFED, BFDE, CEFD are all a parallelogram (opposite sides are parallel). So, $\angle A\ = \angle E$ , $\angle F\ = \angle C$ , $\angle B\ = \angle D$ . So DEF and ABC are equiangular. So, two triangles are similar. A simple, elementary proof :)

In 'fancier' terms, it is a spiral similarity about centroid with ratio $=-\frac 12$ ;).