The Largest Rectangle Inscribed in Any Triangle

From Mathcounts Mini : Maximum area of inscribed rectangles and triangles

$\mathrm{\Delta }EHI\sim \mathrm{\Delta }EFG$ $\to$ ${\displaystyle \frac{a}{c}}={\displaystyle \frac{d-b}{d}}$$\to$ $a={\displaystyle \frac{c(d-b)}{d}}={\displaystyle \frac{-c(b-d)}{d}}$

We are going to find out what the largest area of a rectangle is with the side length a and b.
It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.

$a\times b={\displaystyle \frac{-c(b-d)\times b}{d}}={\displaystyle \frac{-c(b^2-bd)}{d}}={\displaystyle \frac{-c{(b-{\displaystyle \frac{1}{2}}d)}^2+{\displaystyle \frac{1}{4}}dc}{d}}$.

From there, you know that when $b={\displaystyle \frac{1}{2}}d$, it will give you the largest area, which is ${\displaystyle \frac{1}{4}}dc$.

$a={\displaystyle \frac{-c(b-d)}{d}}={\displaystyle \frac{-c({\displaystyle \frac{1}{2}}d-d)}{d}}={\displaystyle \frac{c(d-{\displaystyle \frac{1}{2}}d)}{d}}={\displaystyle \frac{1}{2}}c$.

Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.

Proof without words from Mr. Rusczyk 

Try using different types of triangles to experiment and see for yourself.
Paper folding is fun !!!!!
It's very cool :D