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 $T$.

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.

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

$\ds AF$

$=$

$\ds FH$

$\ds HG$

$=$

$\ds GB$

$\ds DF$

$=$

$\ds \frac 1 2 GH$

$\ds \triangle CDE$

$\cong$

$\ds \triangle HDE$

$\ds \triangle ADF$

$\cong$

$\ds \triangle HDF$

$\ds \triangle BGE$

$\cong$

$\ds \triangle HGE$

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 $T$.

$\blacksquare$