2
$\begingroup$

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.

enter image description here

GeoGebra applet that demonstrates this proposition can be found here.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.