Problem 10 from Chapter 3 wants us to show that if we connect the midpoints of the sides of an equilateral triangle, the resulting figure has four smaller congruent equilateral triangles, like in the figure below:

That is, we want to show that $$\triangle AFD \cong \triangle BEF \cong \triangle CDE \cong \triangle DEF.$$

Note (i) $\dfrac{AD}{AC} = \dfrac{1}{2}$ since $D$ is the midpoint of $AC$, (ii) $\angle DAF = \angle CEB$, since it is the same angle, and (iii) $\dfrac{AF}{AB} = \dfrac{1}{2}$ since $F$ is the midpoint of $AB$. Thus by SAS $\triangle AFD \sim \triangle ABC$. We can do the same to show that $\triangle BEF$ and $\triangle CDE$ are also similar to $\triangle ABC$, and since $AF=BE = CD$, we have that these three triangles are also congruent to each other.

Then by SSS $\triangle DEF$ is also congruent to these three triangles.

Therefore all four triangles are congruent equilateral triangles.