Show that in every regular pentagon when two diagonals intersect, one of the obtained segments is equal to the side of the pentagon.
I lost myself in my solution, to be honest. Firstly, we can calculate the angle of the regular pentagon: $\measuredangle A = \dfrac{5-3}{5}\cdot180^\circ=3\cdot36^\circ=108^\circ$. On my diagram the diagonals are $AD$ and $BE$. They intersect at $P$. It seems that we should show that $PBCD$ is a parallelogram (but since $BC=CD=a$ it will become a rhombus; so then aren't two of the obtained segments equal to the side of the pentagon??). What's the most straightforward path to the solution? I calculated $\measuredangle ABE=\measuredangle AEB=\dfrac{180^\circ-108^\circ}{2}=36^\circ$. Triangles $ABE$ and $EAD$ are congruent since $AB=AE=ED$ and $\measuredangle A=\measuredangle E$. This means $\measuredangle EAD=\measuredangle ADE=36^\circ$ and $BE=AD$.
Best Answer
$\def\deg{^\circ}$
$\angle ABC = 108\deg$ and $\angle ABE = 36\deg$, so $\angle EBC = 72\deg$.
Then $\angle EBC + \angle BCD = 72\deg + 108\deg = 180 \deg$, so $BE\parallel CD$. Similarly $AD \parallel BC$.
So $PBCD$ is a parallelogram.