# Further properties of Van Aubel Configuration

### Problem

Let $ABC$ be a triangle, $M_a$ the midpoint of $BC,$ $H$ the orthocenter of $\Delta ABC,$ $V_b,V_c,V'_b,V'_c$ the centers of the four squares, as in the diagram below:

Show that:

1. Six points $V_b,V'_b,M_a,H_a,V_c,V'_c$ lie on a circle.

2. $V_bV_c$ is a diameter of this circle.

3. Reflection $A'$ of $A$ in center of this circle is a fixed point, independent of $A.$

Note that this statement nicely complements van Aubel's theorem.

Let the two squares on the sides of $\Delta ABC$ be $ABGF$ and ACDE.$Let$I$be the midpoint of$EF.$By the Finsler-Hadwiger theorem$M_{a}V_{b}IV_{c}$is a square, making the fur points$M_{a},V_{b},I,V_{c}$concyclic, with$V_{b}V_{c}$a diameter of their circumcircle. Let it be$(O).$Triangles$AEF$and$ABC$are friendly so that the extension of the median$AI$in$\Delta AEF$is an altitude in$\Delta ABC.$In other words,$A$lies on$IH_a.$From$IM_a$being a diameter of$(O)$and$IH_{a}\perp H_{a}M_{a}$it follows that$H_a\in(O).$Redefine$V'_b$to lie on the intersection of the line through$I$parallel to$AC$and the line through$M_a$parallel to$BH_b.$First of all, this causes$\angle IV'_{b}M_a =90^{\circ}$and, therefore,$V'_b\in (O).$It remains to be shown that$V'_b$is the center of the square with side$CH_b.$Now,$M_a$is the midpoint of side$BC$in$\Delta BCH_b$so that$V'_b$is equidistant from$C$and$H_b.$Moreover,$\angle V_{c}V'_{b}I=\angle V_{c}M_{a}I=45^{\circ},$due to the Finsler-Hadwiger theorem. Thus$V'_{b}$is indeed the center of the square with side$CH_b.$Similarly we show that$V_c$also lies on$(O).V_{b}V_c$is the midline in$\Delta ADG$parallel to$DG.$It follows that$AO$meets$DG$at a point at the same distance to$O$as$A,$i.e.$A'.$In addition,$O$being the midpoint of$V_{b}V_c,A'$is the midpoint of$DG.$By Bottema's theorem,$A'$does not depend on$A.$### Acknowledgment The statement has been posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page. There are three addional solutions posted at the artofproblemsolving.org site. ### Extra Points$V_c,A',H_b,V'_b\$ are collinear.

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