Notes
two overlapping triangles solution

Solution to the Two Overlapping Triangles Puzzle

Solution by Similar Triangles

In the above diagram, $D$ is the midpoint of the side of the orange triangle and $F$ is so that angle $A \hat{F} B$ is a right-angle. Since angle $E \hat{A} B$ is a right-angle, angles $E \hat{A} D$ and $D \hat{A} B$ add up to a right-angle, and so angles $A \hat{B} F$ and $E \hat{A} D$ are equal. This implies that triangles $E D A$ and $A F B$ are similar. From this, the ratio of lengths $A D : A E$ is the same as $B F : B A$. Since $B A$ is half the side length of the equilateral triangle of which $A E$ is the height, $B F$ has length half the side length of an equilateral triangle of which $A D$ is the height. Therefore $B F$ and $C D$ have the same length and so $B C D F$ is a rectangle. Since angle $D \hat{C} A$ is $60^\circ$, this leave angle $A \hat{C} B$ as $30^\circ$.

Solution by Angles in the same segment and Angle at the centre is twice the angle at the circumference

In the above diagram, point $O$ is the midpoint of the side $B E$, so the circle centred at $O$ that passes through $B$ also passes through $E$. As $A$ is the midpoint of the base of the yellow equilateral triangle, $O A$ is the same length as $O B$ and so the circle also passes through $A$. Angle $E \hat{C} A$ is $60^\circ$ which is the same as angle $E \hat{B} A$, so by the result that angles in the same segment are equal, $C$ also lies on the circle. Therefore since the angle at the centre is twice the angle at the circumference, angle $A \hat{C} B$ is half of angle $A \hat{O} B$ which is $60^\circ$. Hence angle $A \hat{C} B = 30^\circ$.

Solution by Transformations

For this solution, view the orange triangle as fixed and the yellow triangle as variable. It is determined by the point $E$ which is free to move on the line that extends the top side of the orange triangle. The point $B$ is obtained from $E$ by a rotation $90^\circ$ clockwise about the point $A$ (which is fixed relative to the orange triangle) and scaling by $\frac{1}{\sqrt{3}}$. Therefore, $B$ moves on a straight line (relative to the orange triangle). To see where this straight line is, consider two special cases.

In this case, $E$ is at a point so that $A B$ lies along $A C$. This is also characterised by the fact that angle $E \hat{A} C = 90^\circ$. In this case, $B$ coincides with $C$.

The second special case is where $E$ is at $D$. In this diagram, angle $D \hat{C} B$ is evidently a right-angle.

So the line that $B$ moves on passes through $C$ at right-angles to the line $D C$. Since angle $D \hat{C} A = 60^\circ$, this shows that angle $A \hat{C} B = 30^\circ$.

Solution by Invariance Principle

The diagram for the second special case above can also be used to demonstrate the solution using the strong invariance principle.