Are the 4 triangles formed by midpoints of of a triangle congruent? I recently purchased Paul Lockhart's math book titled "Measurement". In this book he asks math questions but avoids giving answers in order to motivate the reader to try the question themselves. I am a highschool freshman and inexperienced mathematician. Please don't judge. 
This was a question that I attempted:
If the midpoints of ANY triangles sides are connected, this will make four different triangles. Are these four triangles congruent? Prove why or why not.
Proof
A(tri)/4 = bh/8       * let's assume that the triangles are congruent
A(par) = 2(tri)       * since ANY two congruent triangles can make a parallelogram
A(par)/8 = bh/8
A(tri)/4 = A(par)/8
8(1/2bh) = 4bh
4bh = 4bh
1 = 1 
 A: Why don't you just use Thale's theorem to show they have the same side lengths?
A: Here's how I'd argue it. Consider the triangle below, where $D$, $E$ and $F$ are midpoints. We can say that $EF$ and $AB$ are parallel and by extension, each interior edge between two midpoints is parallel to the exterior edge with the third midpoint. So then the triangle can be split into the four parallelograms $AFED$, $DFCE$ and $DFEB$. So then each pair of the $3$ triangles, except the middle one, share $2$ edges on opposite sides of a parallelogram and their third edge on an edge of $ABC$. For example, for $FEC$ and $ADF$, the edges $FE$ and $AD$ are opposite each other on a parallelogram and so are the edges $DF$ and $EC$. Since $AF=FC$, $AFD\cong FCE$. The inside triangle has all its edges on opposite sides of a parallelogram so all $4$ small triangles must be congruent.

A: It can be proved easily by using mid point theorem.  If $D$ and $E$ are mid points of sides $AB$ and $AC$ respectively, $DE$ is parallel to side $BC$ and Half in length to $BC$.  Also, $\angle D = \angle B$ and $\angle E = \angle C$, therefore, $\Delta ADE$ and $\Delta ABC$ are similar.  We know that if two similar triangles sides are in ratio $a : b$ then the ratio of their areas = ${a^2}:{b^2}$.  So their areas in the ratio ${1^2}:{2^2} = 1:4$

Similarly we can prove that $\Delta BDF$ and $\Delta ABC$ = ${1^2}:{2^2} = 1:4$
Also, $\Delta CEF$ and $\Delta ABC$ = ${1^2}:{2^2} = 1:4$
So it is clear that, $\Delta ADE = \Delta BDF = \Delta CEF = 1$
Therefore, $\Delta DEF$ = $ 4 - 3 = 1$
