Notes
triangle in a rectangle solution

Solution to the Triangle in a Rectangle Puzzle

Solution by Angles in parallel lines, Angles in a Triangle

Using alternate angles, angle $E \hat{B} A$ is equal to angle $B \hat{E} C$ and so triangle $A B E$ is isosceles. Angles $B \hat{A} F$ and $F \hat{A} D$ add up to $90^\circ$, so since angles $F \hat{A} D$ and $F \hat{B} A$ are equal, angles $B \hat{A} F$ and $F \hat{B} A$ add up to $90^\circ$ meaning that angle $A \hat{F} B$ is $90^\circ$ since angles in a triangle add up to $180^\circ$.

Since triangle $A B E$ is an isosceles triangle and angle $A \hat{F} B$ is a right-angle, $F$ is the midpoint of $E B$. This means that the height of $F$ above $A B$ is one half of the height of $E$. This establishes the shaded area as one quarter of the area of the total rectangle.

Solution by Invariance principle

One way to see the variation in this puzzle is to start with the angle $A \hat{E} C$. The rest of the diagram can be constructed starting with this angle.

Fix a horizontal line and a point $E$ on it. The point $C$ will also lie on this line and to the right of $E$, so it can be used to label the angle $A \hat{E} C$ unambiguously even before its location is fully determined. Choose a point $A$ below the line and to the left of $E$ so that angle $A \hat{E} C$ is obtuse. Let $D$ be on the line so that angle $A \hat{D} E$ is a right-angle. The point $B$ lies on the angle bisector of angle $A \hat{E} C$ so that $A B$ is parallel to $E C$. Then $C$ is so that angle $E \hat{C} B$ is a right-angle. Finally, $F$ is the point on $E B$ such that angle $D \hat{A} F$ is half of angle $A \hat{E} C$.

Two special cases that make the answer simpler to see are when angle $A \hat{E} C = 90^\circ$ and when it is $120^\circ$.

When angle $A \hat{E} C = 90^\circ$, the highlighted angle is $45^\circ$ and the rectangle is a square. The highlighted area is one quarter of that square.

When angle $A \hat{E} C = 120^\circ$, the highlighted angle is $60^\circ$ and the inner triangle is equilateral. The four smaller triangles (of which the shaded triangle is one) are all congruent $30^\circ - 60^\circ - 90^\circ$ triangles and so the shaded area is one quarter of the total area.