Find the side of the pentagon

Question

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 $6$.

Answer 1

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.

Answer 2

hint: let $AB = CD = a, BC = 2a.$ you can use ptolemys theorem on the quad $ABCE$ to compute $AC = 5a/2.$ from here you can compute all the angles of the triangle $ABC$ now, use the triangle $ACE$ to compute $AC$ which in turn gives you $a.$ remember $\angle AEC + \angle ABC = 180^\circ.$