Problem 11.4. Given a regular pentagon $A B C D E$. On side $A E$ there is a point $K$, and on side $C D$ there is a point $L$. It is known that $\angle L A E + \angle K C D = 108^{\circ}$, and $A K: K E = 3: 7$. Find $C L: A B$.

A regular pentagon is a pentagon in which all sides are equal and all angles are equal.

#
# Answer: 0.7.

Solution. Let's start by formulating the known facts that we will use in the solution.

- The sum of the angles of an $n$-sided polygon is $180^{\circ}(n-2)$. In particular, each angle of a regular pentagon is $\frac{540^{\circ}}{5}=108^{\circ}$.
- All five diagonals of a regular pentagon are equal. Each diagonal also cuts off an isosceles triangle from the pentagon with a vertex angle of $108^{\circ}$ and base angles of $36^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2fdcf97aa9799d0d4cd6g-43.jpg?height=385&width=398&top_left_y=966&top_left_x=528)

Fig. 14: to the solution of problem 11.4

Let $\angle L A E = \alpha$, then $\angle K C D = 108^{\circ} - \alpha$ and $\angle B C K = \alpha$ (Fig. 14). Note that

\[
\begin{aligned}
\angle C K A & = 360^{\circ} - \angle K A B - \angle A B C - \angle B C K = 360^{\circ} - 108^{\circ} - 108^{\circ} - \alpha = \\
& = 360^{\circ} - \angle L D E - \angle D E A - \angle E A L = \angle A L D.
\end{aligned}
\]

We will prove that triangles $A L D$ and $C K A$ are equal. We already have the equality of one pair of angles. Next, $\angle A D L = \angle C D E - \angle A D E = 108^{\circ} - 36^{\circ} = 72^{\circ}$; similarly, $\angle C A K = 72^{\circ}$, as this is the analogous angle between the diagonal and the side of the pentagon.

Then the third angles of the triangles are also equal. Therefore, triangles $A L D$ and $C K A$ are equal by the side $A D = A C$ and the adjacent angles; hence, $L D = A K$.

Denoting the length of $A K$ as $3x$, we have $K E = 7x$ from the condition and $A E = A K + K E = 10x$. Moreover, $C D = A B = A E$, since these are the sides of the pentagon. We get that $C L = C D - L D = 7x$. Finally, $C L : A B = 7x : 10x = 0.7$.

Remark. The equality of the triangles can be established by considering a rotation of $144^{\circ}$ around the center of the pentagon, which maps point $C$ to point $A$. From the equality of the angles given in the condition, it follows that angle $B C K$ will map to $E A L$ and segment $C K$ to segment $A L$; then the triangles will coincide.