To edit the convex, equilateral $A B C D E$ pentagon, given that the diagonals $A C$ and $A D$ are provided, and it is known that $\angle B A E = 2 \cdot \angle C A D$.

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To edit the convex, equilateral $A B C D E$ pentagon, given that the diagonals $A C$ and $A D$ are provided, and it is known that $\angle B A E = 2 \cdot \angle C A D$.
Due to convexity, the rays $AC$ and $AD$ pass through the convex $BAE$ angle domain, and due to the angle condition, $\angle BAC + \angle EAD = \angle CAD$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a5cb9b6ba051fb23337ag-1.jpg?height=463&width=469&top_left_y=278&top_left_x=817)

Therefore, and because $AB = AE$, if we reflect the triangle $ABC$ around $AC$ and the triangle $AED$ around $AD$ onto the triangle $ACD$, the new positions of $B$ and $E$ coincide. This point $O$ is as far from $C$ and $D$ as it is from $A$, since $OA$, $OC$, and $OD$ are each the reflection of some side and thus equal. Therefore, $O$ is the center of the circumcircle of triangle $ACD$. Furthermore, the length of side $CD$ is the same, so $OCD$ is an equilateral triangle, and thus

$$
\angle CAD = \frac{1}{2} \angle COD = 30^\circ.
$$

This angle and the given diagonals determine the triangle $ACD = H$. Finally, by reflecting $O$ over $AC$ and $AD$, we obtain the vertices $B$ and $E$.

The correctness of the construction is evident. $H$ is formed regardless of the size ratio of the diagonals, and thus $B$ and $E$ can also be constructed. However, the solution is only valid if the resulting $O$ is inside $H$. This is true if $\angle ACD$ and $\angle ADC$ are greater than $60^\circ$, and thus the larger of the two is an acute angle. This means that the smaller of $AC$ and $AD$ is greater than the other's perpendicular projection onto the first line, or its $\sqrt{3}/2$ multiple. This can be written in the following form:

$$
\frac{\sqrt{3}}{2} < \frac{AC}{AD} < \frac{2}{\sqrt{3}}
$$

Gáspár Gyula (Miskolc, Herman O. Gymn., I. o. t.)

Note. Based on the coincidence of the reflections of $B$ and $E$ at $O$, we can proceed as follows: $ABCO$ and $AEDO$ are clearly rhombuses, so $BC \parallel AO \parallel ED = CD$, and $BCDE$ is a rhombus, $BE = CD$, thus $ABE$ is an equilateral triangle, $\angle BAE = 60^\circ$ and $\angle CAD = 30^\circ$.

Schmidt Ferenc (Miskolc, Földes F. Gymn., II. o. t.)