Triangle to Rectangle

Let the triangle be $ABC$ , with $BC$ being the longest side. Let $D$ be the midpoint of $AB$ and $E$ be the midpoint of $AC$ . Let $F$ be the foot of the altitude from $A$ to $DE$ . Then, because $DE \parallel BC$ by Midpoint Theorem, $\angle ADF = \angle ABC$ , so $\angle DAF + \angle ABC = 90^{\circ}$ . Similarly, $\angle EAF + \angle ACB = 90^{\circ}$ . If we rotate triangle $ADF$ $180^{\circ}$ about $D$ and rotate triangle $AEF$ $180^{\circ}$ about $E$ , we get that $B$ and $A$ coincide in $ADF$ (since $D$ is the midpoint of $AB$ ), and $A$ and $C$ coincide in $AEF$ . Thus, we must get a quadrilateral with 4 right angles, so it must be a rectangle. Therefore, we cut the triangle along a medial line and cut along the altitude of the smaller triangle formed.