Largest Parallelogram Contained in Triangle

Theorem
Let $T$ be a triangle.

Let $P$ be a parallelogram contained within $T$.

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

Then:


 * $(1): \quad$ One side of $P$ 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 $P$ bisect the other two sides of $T$


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

Proof
We will first find the maximum area of $P$ when $(1)$ is satisfied, that is, when $P$ is inscribed in $T$.

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


 * Largest-parallelogram-in-triangle-1.png

Here $DEGF$ is our inscribed parallelogram $P$.

Since $FG \parallel BC$, by Equiangular Triangles are Similar:
 * $\triangle AFG \sim \triangle ABC$

Let $FG : BC = 1 : r = \text {Height of } \triangle AFG : \text {Height of } \triangle ABC$.

The area of $T$, which is fixed, is given by:
 * $\dfrac {BC \times \text {Height of } \triangle ABC} 2$

The area of $P$ is given by:

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

Notice that:

Equality holds $r = 2$ for the first inequality.

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

This implies $AF : AB = AG : AC = 1 : 2$.

Hence both sides $AB$ and $AC$ are both bisected by vertices of $P$.

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

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

Proof of $(1)$
Now we consider the cases where $P$ is not inscribed in $T$.

Suppose less than $3$ vertices of $P$ lie on the sides of $T$.

By constructing parallel lines to the sides of $T$, we can find a smaller triangle $T'$ that is similar to $T$, and the sides of $T'$ touches at least $3$ vertices of $P$.


 * Largest-parallelogram-in-triangle-2.png

In the above figure, $\triangle A'B'C'$ is our $T'$.

The case where all $4$ vertices of $P$ lie on the sides of $T'$ has been covered above.

Suppose, then, that only $3$ vertices of $P$ lie on the sides of $T'$.

We can split $T'$ using a line parallel to one of the sides of $P$ that passes through a vertex of $T'$.

We connect the base of that line to the vertex of $P$ that does not lie on the sides of $T'$.


 * Largest-parallelogram-in-triangle-3.png

In the figure above, we can see that $P$ has also been split into two parallelograms, both are inscribed in the two smaller triangles $\triangle XB'C'$ and $\triangle XYC'$.

By our results above:

so the area of $P$ cannot exceed half the area of $T$.

Hence we see that half the area of $T$ is indeed the maximum area of $P$, and occurs when all three conditions above are satisfied.