Notes
rectangle across a rectangle solution

Solution to the Rectangle Across a Rectangle Puzzle

Solution by Angle at the Circumference is Half the Angle at the Centre

Let $x$ be the length of the diagonal $A C$. Applying Pythagoras' theorem to the right-angled triangle $A E C$ shows that $x$ satisfies $x^2 = 7^2 + 1^2 = 50$. Now let $y$ be the length of $B C$, so the height of rectangle $A B C D$. Applying Pythagoras' theorem to $A B C$ shows that $x^2 = y^2 + 5^2$, so $y^2 = 25$ and $y = 5$. As this is the same length as $A B$, the rectangle $A B C D$ is actually a square.

Now, as in the above diagram, consider the circumcircle of the square $A B C D$, with the point labelled $O$ at its centre. This also passes through the vertices of the other rectangle as the points $A$ and $C$ are on it. Since the angle at the circumference is half the angle at the centre, angle $A \hat{E} C$ is half of angle $A \hat{O} C$. This latter angle is $90^\circ$, hence $A \hat{E} C = 45^\circ$.