Largest Rectangle Contained in Triangle

Theorem
Let $T$ be a triangle.

Let $R$ be a rectangle contained within $T$.

Let $R$ have the largest area possible for the conditions given.

Then:


 * $(1): \quad$ One side of $R$ is coincident with part of one side of $T$, and hence two vertices lie on that side of $T$


 * $(2): \quad$ The other two vertices of $R$ bisect the other two sides of $T$


 * $(3): \quad$ The area of $R$ is equal to half the area of $T$.

Proof
Note that a rectangle is a parallelogram.

By Largest Parallelogram Contained in Triangle, the area of $R$ cannot exceed half the area of $T$.

Hence we only need to show that when the first two conditions above are satisfied, the area of $R$ is exactly half the area of $T$.

Consider the diagram below.


 * Largest-rectangle-in-triangle.png

Since $AD = DC$ and $CE = EB$:

and so the area of $R$ is equal to the area of the parts of $T$ not included in $R$.

That is, the area of $R$ is exactly half the area of $T$.