How do I prove both arcs are equal?

Question

As in the following image, the segments AD, DB, BE and EC make the same angle (x) relative to the diameter of the circle QP. How can I prove the arcs L1 (AB) and L2 (BC) are equal?

Answer 1

This question is trivially answered by simply extending $AD$ and $CE$ and observing their intersection will meet at the circle at a point $B'$, for the reason that we can simply reflect $A, B, C$ across diameter $QP$ to $A', B', C'$ and the given angles force $D$ to be collinear with $AB'$ and $E$ collinear with $CB'$. Thus $\angle AB'C$ is an inscribed angle, and $\triangle DB'E$ is is isosceles; therefore, $\angle AB'B = \angle BB'C$ and by the inscribed angle theorem, the subtended arcs $L_1$ and $L_2$ are equal.

Answer 2

Just an illustration of @heropup's answer.