[alogo] Bisector cross

Consider triangle ABC and its circumcircle c. For a point A* moving on c, construct triangle A*BC and the bisectors AF and A*G of the two triangles. Consider then the circles inscribed and escribed in angles A and A*. Their centers form a rectangle EFGH (see BisectorRectangle.html ). Analogous constructions can be done for the pairs of triangles (BCA* , ACA*), ( BAA*, CAA*) and (BAA*, BAC). In total this gives four rectangles forming a cross as shown.

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]
[2_0] [2_1] [2_2]

A clue fact is that points B, G and J are on a line. In fact, angle(GBC) is half the angle(GIC) = angle(A*AC). Hence BG must pass through the middle J of arc A*C. Then angle(FGB) is half of the angle(BIF)=angle(BIA), which is double the angle(JGM). Thus, F, G, M are on a line, completing the proof for the beautiful cross.

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