# Solution to the Triangle in a Rectangle Puzzle +-- {.image} [[TriangleinaRectangle.png:pic]] > The three marked angles are equal. What fraction of the rectangle is shaded? =-- ## Solution by [[Angles in parallel lines]], [[Angles in a Triangle]] +-- {.image} [[TriangleinaRectangleLabelled.png:pic]] =-- 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 $120^\circ$. +-- {.image} [[TriangleinaRectangle90.png:pic]] =-- 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. +-- {.image} [[TriangleinaRectangle120.png:pic]] =-- 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.