## Task 20/85

Let $A B C D E$ be a regular pentagon and $F$ the intersection of the diagonals $A C$ and $B D$. Calculate the ratio $D F: B F$.
We use the auxiliary theorem: "The diagonals emanating from a vertex of a regular pentagon trisect the pentagon angle."

Proof of the auxiliary theorem:

Let $A B C D E$ be a regular pentagon. Since $B C=C D=D E$, the angles $B A C, C A D, D A E$ are peripheral angles of the pentagon's circumcircle over equal chords. Consequently, according to the peripheral angle theorem,

$$
\measuredangle B A C=\measuredangle C A D=\measuredangle D A E
$$

Since $\measuredangle B A C+\measuredangle C A D+\measuredangle D A E=\measuredangle B A E$,

$$
\measuredangle B A C=\measuredangle C A D=\measuredangle D A E=\frac{1}{3} \measuredangle B A E=36^{\circ}
$$

By cyclic permutation, the claim follows for all vertices. According to the auxiliary theorem, $AF$ is the bisector of the angle $B A D$ in the triangle $A B D$. According to a well-known theorem, the angle bisector divides the opposite side in the ratio of the two adjacent sides. Therefore, $D F: B F=A D: A B$. According to the sine rule, however,

$$
A D: A B=\sin \measuredangle D B A: \sin \measuredangle B D A=\sin 72^{\circ}: \sin 36^{\circ}=2 \cos 36^{\circ}
$$

because $\sin 72^{\circ}=2 \sin 36^{\circ} \cos 36^{\circ}$. Thus, $D F: B F=2 \cos 36^{\circ} \approx 1.618$.