# Solution to the Semi-Circle Between Two Quarter Circles Puzzle +-- {.image} [[SemiCircleBetweenTwoQuarterCircles.png:pic]] > What's the angle? =-- ## Solution by [[Pythagoras' Theorem]], [[Isosceles Triangle]], [[Angles in a Triangle]], and [[Angles at a Point on a Straight Line]] +-- {.image} [[SemiCircleBetweenTwoQuarterCirclesLabelled.png:pic]] =-- 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 triangle|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: $$ 180^\circ - \left(90^\circ - \frac{a}{2}\right) - \left(90^\circ - \frac{b}{2}\right) = \frac{a + b}{2} = 45^\circ $$