The figure illustrates a regular polygon with 9 sides. The measure of the side of the polygon is \( a \), the measure of the smallest diagonal is \( b \), and the measure of the largest diagonal is \( d \).

![](https://cdn.mathpix.com/cropped/2024_05_01_64d7a736a938f072abd8g-6.jpg?height=400&width=388&top_left_y=497&top_left_x=1005)

Figure 58.1

(a) Determine the measure of angle \( \angle BAE \).

(b) Show that \( d = a + b \).

![](https://cdn.mathpix.com/cropped/2024_05_01_64d7a736a938f072abd8g-6.jpg?height=409&width=511&top_left_y=1149&top_left_x=133)

Figure 58.2

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(a) The measure of the internal angle of a regular enneagon (9 sides) is equal to $180^{\circ} \times 7 / 9 = 140^{\circ}$.

Consider now the pentagon $A B C D E$, as indicated in the figure. The sum of its internal angles is $180^{\circ}(5-2)=540^{\circ}$. We know that $A \hat{B C}=\hat{B C D}=\hat{C D E}=140^{\circ}$ and by the symmetry of the figure, we know that $E \hat{A} B=A \hat{E} D=\alpha$. Therefore,

$$
2 \alpha + 3 \times 140^{\circ} = 540^{\circ}
$$

from which $\alpha=60^{\circ}$.

(b) Let $X$ be the point of intersection of the lines $A B$ and $D E$. Since $X \hat{A} E = X \hat{E} A = 60^{\circ}$, the triangle $A X E$ is equilateral. The triangle $B X D$ is also equilateral, as the line $A E$ is parallel to the line $B D$.

Thus, we have $A X = A E$ and $B X = B D$. Since $A X = A B + B X$, we have $A E = A B + B D$, that is, $d = a + b$.