Notes
a square in a square solution

Solution to the A Square in a Square Puzzle

Solution by Area of a Triangle and Angles in an Equilateral Triangle

In the above diagram, $D$ is the point where $D C$ and $C B$ have the same length, and $E$ is so that $D E$ is perpendicular to $A B$.

Label the lengths in triangle $A C B$ by $a$, $b$, $c$ where $a$ is the length of $C B$, $b$ of $A C$, and $c$ of $A B$. Let $h$ be the length of $D E$.

Triangle $A B C$ has area $\frac{1}{2} a b$, and four of them have half the area of the outer square, so $4 \times \frac{1}{2} a b = \frac{1}{2} c^2$, and thus $a b = \frac{1}{4} c^2$.

Triangle $A D B$ has twice the area of triangle $A C B$, so has area $a b$. Its area is also $\frac{1}{2} c h$. Putting these together shows that $h = \frac{1}{2} c$.

Since $A D$ is congruent to $A B$, triangle $A D E$ is a right-angled triangle with $D E$ half the length of $A D$, hence it is half of an equilateral triangle and so angle $D \hat{A} E$ is $30^\circ$, meaning that angle $C \hat{A} B$ is $15^\circ$.

Solution by Trigonometry and the Double Angle Formula for Sine

With the lengths as above, let $x$ be angle $C \hat{A} B$. Then:

As above, $a b = \frac{1}{4} c^2$ so:

Therefore $2 x = 30^\circ$ and so $x = 15^\circ$.