Notes
two triangles in a rectangle solution

Solution to the Two Triangles in a Rectangle Puzzle

Solution by Similar Triangles and Area of a Triangle

With the diagram labelled as above, let $a$ and $b$ be the lengths of the sides of the outer rectangle, with $a$ the length of $A C$ and $b$ the length of $A G$.

Then $D I$ has length $a - b$, so the area of the orange triangle, $C I E$, is $\frac{1}{2} b ( a - b)$.

Let $c$ be the length of $B I$, so $F I$ has length $b - c$. The area of the yellow triangle is then $\frac{1}{2} a (b - c)$.

Triangle $A B I$ is similar to triangle $A C E$ since both share angle $I \hat{A} B$ and both are right-angled triangles. Therefore, $\frac{c}{b} = \frac{b}{a}$ which rearranges to $a c = b^2$. This means that $a (b - c) = a b - a c = a b - b^2 = b(a - b)$.

Hence the area of the yellow triangle is $\frac{1}{2} a (b - c) = \frac{1}{2} b (a - b)$ which is the area of the orange triangle.

Solution by Area of a Triangle

With the same diagram as above, consider the triangles $G J I$ and $I J C$. To work out their areas, we need the perpendicular height of $I$ above $G C$ and the lengths of $G J$ and $J C$. Since $J$ is the midpoint of $G C$, these lengths will be the same and so the areas of $G J I$ and $I J C$ are the same.

Then triangles $G J E$ and $E J C$ also have the same area since each is a quarter of the full rectangle. Therefore, the orange and yellow areas are the same.

(A similar result follows by considering triangles $A G I$ and $A C I$ and noting that the perpendicular height of $C$ above $A I$ is the same as that of $G$.)