Largest Rectangle Contained in Triangle

Theorem
Let $T$ be a triangle

Let $R$ be a rectangle inscribed 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
It is intuitively obvious that the largest rectangle in $T$ has one side coincident with part of one side of $T$.


 * Largest-rectangle-in-triangle.png

That area of $R$ is equal to half the area of $T$ is proved by noting that:


 * $\triangle CDE = \triangle HDE$


 * $\triangle ADF = \triangle HDF$


 * $\triangle BGE = \triangle HGE$

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