# Solution to the Pentagon with two Semi-Circles Puzzle +-- {.image} [[PentagonwithtwoSemiCircles.png:pic]] > The regular pentagon is tangent to both semicircles at the point where the three meet. What's the angle between the bases of the semicircles? What if I started with other regular polygons? =-- ## Solution by [[Angle Between a Radius and Tangent]] and [[Angles in a Regular Pentagon]] +-- {.image} [[PentagonwithtwoSemiCirclesLabelled.png:pic]] =-- In the above diagram, the points labelled $B$ and $H$ are the centres of their semi-circles. Since the [[angle between a radius and tangent]] is $90^\circ$, angles $B \hat{E} D$ and $F \hat{E} H$ are both $90^\circ$. Angle $D \hat{E} F$ is the [[interior angle]] in a [[regular pentagon]] so is $108^\circ$. Since the [[angles at a point]] add up to $360^\circ$, this leaves $72^\circ$ for angle $H \hat{E} B$. Triangles $A B E$ and $A E G$ are both [[isosceles]], so angles $B \hat{A} H$ are $H \hat{E} B$ are equal and therefore the angle between the bases of the semi-circles is $72^\circ$.