IRL ABooklet

Solution 1. Because the arcs $AB$ and $CD$, as well as $BC$ and $DE$ are equal, the cyclic quadrilaterals $ABCD$ and $BCDE$ are isosceles trapeziums with $BC \parallel AD$ and $CD \parallel BE$. Hence, $BFDC$ is a parallelogram. As a consequence we see that $|BF| = |CD| = |BA|$ and so $\triangle AFB$ is isosceles.

Triangles $\triangle EDM$ and $\triangle BCM$ are congruent by SAS, since $|ED| = |BC|$, $\angle EDC = \angle DCB$ in isosceles trapezium, and $|DM| = \frac{1}{2}|CD| = |CM|$. Hence $|EM| = |MB|$.

To complete the solution, we prove that $|MB| = |MN|$, where $N$ is the midpoint of $AF$. Here are three ways doing so.

First way: Let $P$ be the midpoint of $AB$. Since $M$, $P$, $N$ are midpoints, $PN \parallel BF \parallel CD$ and $|CM| = \frac{1}{2}|CD| = \frac{1}{2}|BF| = |NP|$. This implies that $CPNM$ is a parallelogram, in particular $|MN| = |CP|$.

On the other hand, triangles $\triangle CBP$ and $\triangle BCM$ are congruent by SAS: $BC$ common, $\angle C = \angle B$ in isosceles trapezium, and $|CM| = \frac{1}{2}|CD| = \frac{1}{2}|AB| = |BP|$. This implies $|MB| = |CP| = |MN|$.

Because $PM$ is the mid-line of the trapezium $ABCD$, $PM \parallel AD$ and so $\angle FAB = \angle MPB$ and $PNFS$ is a parallelogram where $S$ is the intersection point of $PM$ and $EB$. Hence,
$$\angle MPN = \angle AFB = \angle FAB = \angle MPB$$
where we have used again that $\triangle AFB$ is isosceles. We now conclude that triangles $\triangle MPN$ and $\triangle MPB$ are congruent by SAS: $MP$ common, $\angle MPB = \angle MPN$ and $|PN| = |BP|$ as shown above. This implies $|MN| = |MB|$.

Third way: As $\triangle BAF$ is isosceles and $AD \parallel BC$, we have $BN \perp AF$ and $BN \perp BC$. Extend the line $NM$ to meet line $BC$ at $Q$.

Since $DN \parallel CQ$ and $M$ is midpoint of $CD$, we get $\triangle MDN \cong \triangle MCQ$ (ASA) and hence $M$ is the midpoint of $NQ$ in the right angled triangle $QBN$. This implies $|MN| = |MQ| = |MB|$.

Solution 2
Let $N$ denote the midpoint of $AF$. We have $\angle ECD = \angle EAF$ since both are standing on the arc $DE$. Since the arcs $AB$ and $CD$ are equal, we also have $\angle DEC = \angle FEA$. This shows that $\triangle EDC$ and $\triangle EFA$ are similar.

As a consequence, we have $\frac{|ED|}{|EF|} = \frac{|CD|}{|AF|}$. Because $M$ and $N$ are the midpoints of $CD$ and $AF$, we now get $\frac{|ED|}{|EF|} = \frac{|MD|}{|NF|}$, which implies $\triangle EDM \sim \triangle EFN$ since $\angle EDM = \angle EDC = \angle EFA = \angle EFN$. Hence $\frac{|ED|}{|EF|} = \frac{|EM|}{|EN|}$. From $\angle DEM = \angle FEN$ we obtain $\angle DEF = \angle MEN$, hence $\triangle EDF$ is similar to $\triangle EMN$.
We have seen above that $\angle FEA = \angle DEC$, and we also have $\angle EAF = \angle CEB$ because the arcs on which they are standing, $DE$ and $BC$, are equal. Therefore, $\angle EFD = \angle FEA + \angle EAF = \angle DEC + \angle CEB = \angle DEF$. This means that $\triangle EDF$ is isosceles with $|ED| = |DF|$ and so $\triangle EMN$ is isosceles with $|EM| = |MN|$ as well.