Notes
semi-circle between two quarter circles solution

Solution to the Semi-Circle Between Two Quarter Circles Puzzle

Solution by Pythagoras' Theorem, Isosceles Triangle, Angles in a Triangle, and Angles at a Point on a Straight Line

Joining the centres passes through the points where the part circles meet. The lengths of each then are the sums of the radii, so are $3$, $4$, and $5$. Since $3^2 + 4^2 = 5^2$, this means that triangle $A C B$ is right-angled by Pythagoras' theorem with the right-angle at $C$. So angles $D \hat{A} E$ and $F \hat{B} D$ sum to $90^\circ$. Let $a$ be angle $D \hat{A} E$ and $b$ angle $F \hat{B} D$, so $a + b = 90^\circ$.

As triangle $A E D$ is isosceles and the angles in a triangle sum to $180^\circ$, angle $E \hat{D} A$ is $90^\circ - \frac{a}{2}$. Similarly, angle $B \hat{D} F$ is $90^\circ - \frac{b}{2}$. Since angles at a point on a straight line add up to $180^\circ$, angle $F \hat{D} E$ is therefore given by: