# Six concyclic points

Can you provide a proof for the following proposition:

Proposition. Let $$\triangle ABC$$ be an arbitrary triangle with excenters $$J_A$$,$$J_B$$ and $$J_C$$ . Let $$G$$ be the orthogonal projection of the $$J_B$$ on the extension of the side $$BC$$ , $$H$$ orthogonal projection of the $$J_B$$ on the extension of the side $$AB$$ , $$I$$ orthogonal projection of the $$J_C$$ on the extension of the side $$AC$$ , $$J$$ orthogonal projection of the $$J_C$$ on the extension of the side $$BC$$ , $$K$$ orthogonal projection of the $$J_A$$ on the extension of the side $$AB$$ and $$L$$ orthogonal projection of the $$J_A$$ on the extension of the side $$AC$$ . Now let $$M$$ be the point of intersection of the line segments $$GH$$ and $$J_AJ_B$$ ,$$N$$ point of intersection of the line segments $$GH$$ and $$J_BJ_C$$ , $$O$$ point of intersection of the line segments $$IJ$$ and $$J_BJ_C$$ , $$P$$ point of intersection of the line segments $$IJ$$ and $$J_AJ_C$$ , $$Q$$ point of intersection of the line segments $$LK$$ and $$J_AJ_C$$ and $$R$$ point of intersection of the line segments $$LK$$ and $$J_AJ_B$$ . I claim that the points $$M$$,$$N$$,$$O$$,$$P$$,$$Q$$,$$R$$ lie on a common circle.

GeoGebra applet that demonstrates this proposition can be found here.

Note that $$HA = BK = (a + b - c) / 2$$, $$\angle NAH = \angle QKB = (\pi - \angle A) / 2$$, and $$\angle NHA = \angle QBK = (\pi - \angle B) / 2$$, hence $$\triangle NHA \cong \triangle QBK$$, hence $$NQ \parallel AB$$, hence $$\angle GMC = (\pi - \angle A) / 2 = \angle BKQ = \angle NQR$$, hence $$M, N, Q, R$$ are concyclic. Similarly, $$M, N, P, O$$ are concyclic and $$P, O, Q, R$$ are concyclic. If these three circles are not the same, then line $$MN$$, line $$QR$$, line $$PO$$ would be their radical axes, and they have to be concurrent, a contradiction.