# Rhombus in Circles

The problem below was posted posted at the CutTheKnotMath facebook page by Cõ Gẫng Lên:

Segments $AB$ and $CD$ are equal. Lines $AB$ and $CD$ intersect at $M.$ Form four circumcircles: $(E)=(ACM),$ $(F)=(ADM),$ $(G)=(BDM),$ $(H)=(BCM).$ Prove that points $E,F,G,H$ form a rhombus, with $\angle EFG=\angle AMC$.

(The applet below illustrates the problem.)

Solution Segments $AB$ and $CD$ are equal. Lines $AB$ and $CD$ intersect at $M.$ Form four circumcircles: $(E)=(ACM),$ $(F)=(ADM),$ $(G)=(BDM),$ $(H)=(BCM).$ Prove that points $E,F,G,H$ form a rhombus, with $\angle EFG=\angle AMC$.

### Solution

It pays to produce perpendicular bisectors of $AM,BM,CM,DM.$ This are the lines that form the quadrilateral $EFGH.$ The lines come in parallel pairs with distances between the lines within the pairs equal to, say, $AB/2.$ This why we get a rhombus. The sides of $\angle EFG$ are perpendicular to those of $\angle AMC,$ and this is why the two are equal.

This is a nice simple configuration. The shape or size of rhombus $EFGH$ does not depend on the relative position of the two given segments, but only on their length and the angles they form. If they are orthogonal, the rhobus becomes a square. If the angle between them is $60^{\circ},$ the rhombus consists of two equilateral triangles $EFG$ and $EHG;$ to construct either one only needs to draw three of the four circles. 