Notes
four congruent triangles solution

Solution to the Four Congruent Triangles Puzzle

Solution by Angles in an Isosceles Right-Angled Triangle and Congruent Triangles

Label the diagram as above, in which $C$ is the point on $B E$ such that line segment $B C$ has the same length as $A B$.

In effect, the diagram with triangles $A B J$, $J B C$, $C E G$, and $G E F$ is obtained from the original by swapping the positions of the two middle triangles. So triangles $J B C$ and $C E G$ are also congruent to the original four.

Since $B D E$ is a straight line, and angles $H \hat{D} B$ and $E \hat{D} I$ are equal, they must be $90^\circ$. So the triangles are all right-angled. This means that angles $J \hat{C} B$ and $E \hat{C} G$ add up to $90^\circ$, since angles in a triangle add up to $180^\circ$, and so angle $G \hat{C} J$ is $90^\circ$, since angles at a point on a straight line add up to $180^\circ$.

Then as $J C$ and $C G$ are corresponding sides of congruent triangles, they have the same length, so triangle $J C G$ is an isosceles right-angled triangle. Angle $C \hat{J} G$ is therefore $45^\circ$.

Then as $C J$ and $E I$ are parallel, angle $E \hat{H} G$ is also $45^\circ$, since corresponding angles are equal. Hence angle $G \hat{H} I$ is $135^\circ$.

Solution by the Compound Angle Formulae in Trigonometry

With the points labelled as above, in which $K$ is such that $J K$ is parallel to $B E$ and $K G E$ is a straight line, let the sides of the main triangle be $a$ and $b$, with $a \lt b$. Let $\theta$ denote angle $H \hat{E} I$. Then $\tan(\theta) = \frac{a}{b}$.

Consider triangle $J K G$. Its side lengths are $a + b$ and $b - a$. Let $\phi$ denote angle $H \hat{G} J$, then:

Since both are acute, this means that $\phi = \theta + 45^\circ$. Therefore, angle $H \hat{G} E = 135^\circ - \theta$ and so angle $G \hat{H} I = 135^\circ - \theta + \theta = 135^\circ$.