Polygons (here triangles) similar to each other w.r. to a common vertex (here C) and inscribable in circles. The lines joining homologous vertices pass through the second intersection point of their circumcircles (here point O). The key fact is that triangle (DCA) is similar to (CEG). The rotation-angle involved in the similarity is equal to ang(DCE), by which also line DA is rotated so as to take position EG. Thus the angles at O and C are equal. The argument, demonstrated here for equilateral triangles, is valid also in the general case.

1) Switch to the selection-tool (CTRL + 1)

2) Catch and move A or B.

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