What's the measure of the segment $PD$ in the pentagon below?
Solution 1:
Assuming $PC = 4$
Notice that $BC = PC$ (as $\angle BPC = \angle PBC$). Similarly, $AE = PE = PC$. So, $PCDE$ is a rhombus and $PD \perp CE$. Say they intersect at $M$. Then by Pythagoras in $\triangle PEM$,
$ \displaystyle PD = 2 PM = 2 \sqrt {PE^2 - \frac{CE^2}{4}}$
We know that the diagonals of a convex regular pentagon are in the golden ratio to its sides.
That leads to $CE = \dfrac{1 + \sqrt 5}{2} \cdot PC = 2 (1 + \sqrt 5)$
$ \therefore PD = 2 \sqrt{16 - (6 + 2 \sqrt 5)} = 2 \sqrt {10 - 2 \sqrt 5}$