Six concyclic points

Question

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.

Answer 1

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.

Answer 2

This is actually the Excircles Radical Circle. Its center is X(10)