The triangle $\triangle ABC$ is equilateral, hence $\angle ABC = 60^\circ$. Similarly, the triangles $\triangle BCD, \triangle BDE$ are equilateral, hence, $\angle CBD = \angle DBE = 60^\circ$. Therefore, the angle $\angle ABE = \angle ABC + \angle CBD = \angle DBE = 60^\circ + 60^\circ + 60^\circ = 180^\circ$ is straight and the points A, B, E are collinear. In addition, AB = BC = BD = BE and B is the midpoint of the segment AE. The congruent triangles $\triangle AEF \cong \triangle AEG$ are isosceles with EA = EF = EG (radii of the circle (E)) and the ratio of the shoulder side to the base is $\dfrac{EA}{AC} = \dfrac{EA}{AB} = 2$. The congruent triangles $\triangle AFM \cong \triangle AGM$ are also isosceles with FA = FM = GA = GM (radii of the congruent circles (F), (G)). In addition, they are similar to the isosceles triangles $\triangle AEF \cong \triangle AEG$, having the base angles at the vertex A common. Thus the ratio of the shoulder side to the base is also equal to $\dfrac{AF}{AM} = \dfrac{AG}{AM} = 2$. But since AF = AG = AB (radii of the circle (A)), we get $\dfrac{AB}{AM} = 2$. The fact that the point M lies on the segment AB follows from symmetry of the congruent circles (F), (G) with respect to the line AB.

Peter wrote: Starting with two points A and B, some circles and points are constructed as shown in the figure: the circle with centre A through B the circle with centre B through A the circle with centre C through A the circle with centre D through B the circle with centre E through A the circle with centre F through A the circle with centre G through A (I think the wording is not very rigorous, you should assume intersections from the drawing) Show that $ M$ is the midpoint of $ AB$. Invalid image file A little history: This problem is essentially the Mascheroni construction with compasses only of the midpoint of a segment $ AB$. I had read this solution many years ago in Kostovski's book ( ed. MIR). How can we find the midpoint of $ AB$ using only a ruler(not graded) and a given circle with known center ? I suggest this challenge for you. Babis

Starting with two points A and B, some circles and points are constructed as shown in the figure: the circle with centre A through B the circle with centre B through A the circle with centre C through A the circle with centre D through B the circle with centre E through A the circle with centre F through A the circle with centre G through A Show that $M$ is the midpoint of $AB$.