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 of $(1)$
By definition of inscribed polygon, all the vertices of the inscribed rectangle must lie on the sides of the triangle.

By Pigeonhole Principle, at least one side of the triangle contains two vertices of the rectangle.

There cannot be three vertices in one side or two sides with two vertices as these cases cannot form a rectangle.

Hence exactly one side of the triangle contains two vertices of the rectangle.

Thus this rectangle in $T$ has one side coincident with part of one side of $T$.

Proof of $(2)$
Consider the diagram below.


 * Largest-rectangle-in-triangle.png

Suppose $AD : DC = 1 : r$, where $r \ge 0$.

Note that $DE \parallel FG$.

By Parallel Transversal Theorem:
 * $BE : EC = 1 : r$
 * $DF : CH = 1 : \paren {1 + r}$

Note also $DF \parallel CH \parallel EG$.

By Parallel Transversal Theorem again:
 * $AF : FH = 1 : r$
 * $BG : GH = 1 : r$

Thus:
 * $AB : FG = \paren {AF + FH + HG + GB} : \paren {FH + HG} = \paren {1 + r} : r$

The area of $T$, which is fixed, is given by:
 * $\dfrac {AB \times HC} 2$

The area of $R$ is given by:

which is $\dfrac {2 r} {\paren {1 + r}^2}$ of the area of $T$.

Notice that:

Equality holds $r = 1$ for the first inequality.

Therefore the maximum of $\dfrac {2 r} {\paren {1 + r}^2}$ occurs at $r = 1$.

This implies $AD : DC = CE : EB = 1 : 1$.

Hence both sides $AC$ and $BC$ are bisected by vertices of $R$.

Proof of $(3)$
At $r = 1$:
 * $\dfrac {2 r} {\paren {1 + r}^2} = \dfrac 2 {2^2} = \dfrac 1 2$

therefore the maximum area of $R$ is equal to half the area of $T$.

Alternatively, we can prove this geometrically by noting that, when $AD = DC$ and $CE = EB$:

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