# Solution to the [[Two Semi-Circles In an Equilateral Triangle]] Puzzle +-- {.image} [[TwoSemiCirclesInanEquilateralTriangle.jpeg:pic]] > Two semicircles inside an equilateral triangle. What’s the angle? =-- ## Solution by [[Angle Between a Tangent and Radius]], [[Angle at the Circumference is Half the Angle at the Centre]], and Angles in a [[Triangle]] +-- {.image} [[TwoSemiCirclesInanEquilateralTriangleLabelled.jpeg:pic]] =-- In the diagram above, $O$ and $Q$ are the centres of their respective semi-circles, $P$ is the point of tangency and $F J$ is the [[tangent]] to both circles at $P$. Since the [[angle between a tangent and radius]] is $90^\circ$, the angles in an [[equilateral triangle]] are $60^\circ$, and the [[angles in a triangle]] add up to $180^\circ$, angle $A \hat{Q} E$ is $30^\circ$. Then as the [[angle at the circumference is half the angle at the centre]], angle $D \hat{P} E$ is $15^\circ$. Similarly, angle $G \hat{P} H$ is also $15^\circ$. Triangle $F E P$ is [[isosceles]] as line segments $E F$ and $P F$ are both tangent to the same circle, as is triangle $F G P$. This means that angle $E \hat{P} G$ is $90^\circ$. Therefore, angle $D \hat{P} H$ is $15^\circ + 90^\circ + 15^\circ = 120^\circ$. Angles $D \hat{Q} E$ and $D \hat{P} E$ a