Any point $X$ is taken on the side $BC$ of $\Delta ABC$. Prove that $AX$ is bisected by the straight line joining the midpoints of $AB$ and $AC$.
This problem is trivial when one uses the concept of similar triangles and the midpoint theorem. However, due to its position in my textbook, there is definitely a solution which uses at most congruent triangles, midpoint theorem and parallel lines (and all resulting definitions + theorems eg. parallelograms, etc.)
Unfortunately I have no idea how to go about the problem without similar triangles. I tried to assume $AX$ and the line joining the midpoints as diagonals of a parallelogram, so that their bisection would follow; however that restricts the position of $X$, so it appears to be fruitless.
call the midpoint of $AC$, $Y$.from $y$ draw a line parallel to $AX$ until in intersects $BC$. call the intesection $Z$. $\Delta ZYC$ is congrunt with $\delta AYW$ ($W$ is the intersection of the midpoints of $AB$ and $AC$ with $AX$). Then connect $Z$ to $W$ to get $\Delta ZWY$. This new triangle is congruent with $\Delta ZYC$, because they share one side and also $WY=ZC$ (because of the previous congruence). The angles between the sides are also equal.
Finally, it remains to show that $\Delta XZW$ is congruent with $\Delta AWY$ and the corresponding sides would be equal and therfore, we are done with the proof.