Q:

Find the side of the pentagon

Let $ABCDE$ be a pentagon inscribed in a circle.  Given that $AB=CD$, $BC=2AB$, $AE=1$, $BE=4$, $CE=8$. Find $DE$. I am unable to use the properties of circle in this question.  All I could do was fund the range of $AB$ which is between $3$ and $6$.

A:

Let $AB=CD=a$ and $BC=2a$.
Note that the quadrilateral $ABCD$ is an isosceles trapezoid, so $AC=BD$.
Applying Ptolmey's theorem in $ABCE$,
$$4AC=8a+2a\implies AC=BD=\dfrac{5a}{2}$$
Applying Ptolmey's theorem in $ABCD$,
$$BD\cdot AC=a^2+2a. AD\implies AD=\dfrac{21a}{8}$$
Applying Ptolmey's theorem in $ABDE$,
$$4AD=a.DE+BD \implies DE=8$$
Se that $a$ cancels out in the last equation, so there wasn't any need to find its value to find $DE$. You just need to know in which quadrilateral to apply Ptolmey's theorem.