## Task 2 - 290732

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0517.jpg?height=442&width=442&top_left_y=1935&top_left_x=313)

Given is a regular nonagon \( A B C D E F G H I \).

a) Determine the number of all diagonals of this nonagon!

b) Determine the size of an interior angle of this nonagon!

c) Let \( K \) be the intersection of the diagonals \( A C \) and \( B E \). Determine the size of the angle \( \angle C K E \)!

Hint: A nonagon is called regular if and only if all its sides have the same length and all its interior angles have the same size.
a) From each corner of the nonagon, exactly six diagonals emanate. If one adds these (for each of the nine corners) six counts, one has in the resulting sum $9 \cdot 6 = 54$ each diagonal exactly twice. Therefore, the number of all diagonals of the nonagon is 27.

b) The six diagonals emanating from one corner divide the nonagon into exactly seven triangles. In each of these triangles, the sum of its interior angles is $180^{\circ}$.

If one adds all the interior angles in these triangles, one gets the sum of all the interior angles of the nonagon; this sum is therefore $7 \cdot 180^{\circ}$. Since all nine interior angles of the nonagon are of the same size, each of these interior angles has the size $\frac{7 \cdot 180^{\circ}}{9} = 140^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0518.jpg?height=471&width=466&top_left_y=387&top_left_x=264)

c) From $\angle A B C = \angle B C D = \angle C D E = 140^{\circ}$ (1) and $A B = B C = C D = D E$ it follows by the base angle theorem and the interior angle theorem that

$$
\angle C A B = \angle A C B = \angle D B C = \angle B D C = \angle E C D = \angle C E D = 20^{\circ}
$$

and by the SAS congruence theorem $\triangle B C D = \triangle C D E$, thus $B D = C E$. (3)

If $L$ is the intersection of $B D$ and $C E$, it further follows from (2) by the converse of the base angle theorem that $C L = D L$. From this and (3), it follows that $B L = E L$. (4)

By the vertical angle theorem and the interior angle theorem, and due to (2), $\angle B L E = \angle C L D = 180^{\circ} - 2 \cdot 20^{\circ} = 140^{\circ}$. From this and (4), it follows by the base angle theorem and the interior angle theorem that $\angle B E L = \angle E B L = 20^{\circ}$. (5)

From (1) and (2), it further follows that $\angle K C E = 140^{\circ} - 2 \cdot 20^{\circ} = 100^{\circ}$; from this and (5), it follows by the interior angle theorem that the desired angle measure $\angle C K E = 180^{\circ} - 100^{\circ} - 20^{\circ} = 60^{\circ}$.