IRL_ABooklet_2023

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 $△AFB$ is isosceles.

Triangles $△EDM$ and $△BCM$ 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 $△CBP$ and $△BCM$ 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 $△AFB$ is isosceles. We now conclude that triangles $△MPN$ and $△MPB$ are congruent by SAS: $MP$ common, $\angle MPB=\angle MPN$ and $|PN|=|BP|$ as shown above. This implies $|MN|=|MB|$.

Third way: As $△BAF$ 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 $△MDN\cong △MCQ$ (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 $△EDC$ and $△EFA$ 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 $△EDM\sim △EFN$ 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 $△EDF$ is similar to $△EMN$. 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 $△EDF$ is isosceles with $|ED|=|DF|$ and so $△EMN$ is isosceles with $|EM|=|MN|$ as well.