10.5. In a regular pentagon $A B C D E$, a point $F$ is marked on side $A B$, and a point $G$ is marked on side $B C$ such that $F G=G D$. Find the angle $C D G$, if the angle $F D E$ is $60^{\circ}$.
Answer: $6^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4881516c6f6287f524c8g-3.jpg?height=985&width=939&top_left_y=1118&top_left_x=681)

Solution. The angles of a regular pentagon are each $108^{\circ}, \angle E D A = 36^{\circ}, \angle F D E = 60^{\circ}$. Therefore, $\angle A D F = 24^{\circ}$.

Additional construction: Take point $O$ such that $B O = C O = B C$. Then $\triangle B O C$ is equilateral, $\angle A B O = O C D = 108^{\circ} - 60^{\circ} = 48^{\circ}$. We get that $\triangle A B O = \triangle O C D$, where $\angle B A O = \angle B O A = \angle C O D = \angle C D O = 66^{\circ}, A O = O D$.

Thus, there exists a circle with center at point $O$ and radius $r = O A = O D$. Let this circle intersect side $A B$ at point $K$. Then
$\angle A K O = \angle K A O = 66^{\circ}, \angle K O A = 48^{\circ}$. We obtain that $\angle A D K = A D F = 24^{\circ}$. Therefore, $K = F$. Then $\angle K A D = \angle B A D = \angle B A E - \angle D A E = 72^{\circ}$, and $\angle A F D = 84^{\circ}$, $\angle O F D = 18^{\circ}$.

Let $H$ be the midpoint of $F D$, then $O H$ and $G H$ are medians, altitudes, and angle bisectors of $\triangle F O D$ and $\triangle F G D$ respectively (thus, $G O \perp F D$). We get $\angle F O H = \angle D O H = \angle D O G = 90^{\circ} - 18^{\circ} = 72^{\circ}$.

$\angle G O D + G C D = 180^{\circ} \Rightarrow O G C D$ is inscribed in a circle $\Rightarrow O C D = \angle O G D = 48^{\circ} = \angle H G D$. Then $\angle G D H = 90^{\circ} - 48^{\circ} = 42^{\circ}$. We obtain that $\angle C D G = 108^{\circ} - 60^{\circ} - 42^{\circ} = 6^{\circ}$.

Comment. It is proven that point $O$, the vertex of the equilateral triangle constructed on segment $B C$, is equidistant from vertices $A, F, D$ - 3 points.