4.31. Two diagonals originating from the same vertex of a regular pentagon divide it into three triangles. Find the ratio of the area of the triangle bounded by these two diagonals to the sum of the areas of the other two triangles.
4.31. Let $A B C D E$ be a regular pentagon (Fig. 4.28). In it, the angles at the vertices are $\frac{(5-2) 180^{\circ}}{5}=108^{\circ}$. In $\triangle A B C$, draw $B F \perp A C$; since $\angle B A C=\angle B C A=\frac{1}{2}\left(180^{\circ}-108^{\circ}\right)=36^{\circ}$, then $\angle A C E=108^{\circ}-2 \cdot 36^{\circ}=36^{\circ}, C F=$ $=a \cos 36^{\circ}$ and $A C=2 a \cos 36^{\circ}$. Next, we find

$$
S_{\triangle A B C}=\frac{1}{2} a^{2} \sin 108^{\circ}=\frac{1}{2} a^{2} \sin 72^{\circ}
$$

$S_{\triangle A C E}=\frac{1}{2} A C^{2} \sin \angle A C E=\frac{1}{2}\left(2 a \cos 36^{\circ}\right)^{2} \sin 36^{\circ}=a^{2} \cos 36^{\circ} \sin 72^{\circ}$.

Thus, $\frac{S_{\triangle A C E}}{2 S_{\triangle A B C}}=\frac{a^{2} \cos 36^{\circ} \sin 72^{\circ}}{a^{2} \sin 72^{\circ}}=\cos 36^{\circ} \approx 0.81$.

Answer: $\approx 0.81: 1$.