# Thanos Kalogerakis's Problem in Circle and Square

### Problem

$BC$ is a diameter of circle $\omega;$ $M$ is the midpoint of one of the arcs $\overset{\frown}{BC};$ point $A$ is on the other arc.

Point $D$ is on line $AB,$ $E$ on line $AC$ such that $MD\perp AB$ and $ME\perp AC.$

Prove that $\displaystyle \frac{AB}{MD}+\frac{AC}{ME}=2.$

### Solution 1

First off, $ADME$ is a square because, due to $M$ being the midpoint of the arc subtending $\angle BAC$ (which is right), $\angle DAM=\angle EAM=45^{\circ}$ and, subsequently, since $ADME$ is clearly a rectangle, $\angle AMD=\angle AME=45^{\circ},$ making $ADME$ a square.

It follows that the required equality can be rewritten as $AB+AC=2MD.$ But, in the diagram, $AC=AE+EC=MD+EC,$ and $AB=AD-BD=MD-BD.$ Adding up gives

$AB+AC=(MD+EC)+(MD-BD)=2MD+(EC-BD).$

Thus the problem will be solved if we show that $EC=BD.$ This is indeed so because triangles $BMD$ and $CM$ are right and equal:

($MD=ME$ and $\angle CME=\angle BMD,$ as having two pairs of perpendicular sides.)

### Solution 2

The problem is a particular case of the one discussed earlier:

With the reference to the notations in the latter problem, when $BD$ a diameter and $x=R,$ the problem reduces to proving $\displaystyle \frac{BC}{y}+\frac{CD}{z}=2$ which is the original problem in different notations.

### Acknowledgment

The problem was posted by Thanos Kalogerakis at the Οι Ρομαντικοι της Γεωμετριας (Romantics of Geometry) facebook group. Additional solutions can be found at the link.

Solution 2 is by Thanos Kalogerakis.