67th NMO Selection Tests for JBMO

We are going to treat only the case when $C\in (GD)$ and $C\in (BH)$, all the other cases being similar.

a) Let $M$ and $N$ be the intersection points of the angle bisector of angle $\angle G$ with the sides $BC$ and $AD$, respectively. Consider $I,K$ the intersection points of the angle bisector of angle $\angle H$ with sides $CD$ and $AB$, respectively. Also, put $\{J\}=MN\cap IK$. From triangle $GBC$ we get $\angle G=180^{\circ }-\angle GBC-\angle GCB=\angle B+\angle C-180^{\circ }$, hence $\angle CMJ=90^{\circ }+\frac{1}{2}(\angle B-\angle C)$. Similarly, $\angle CIJ=90^{\circ }+\frac{1}{2}(\angle D-\angle C)$. From the quadrilateral $CMJI$, using the fact that $\angle B+\angle D=180^{\circ }$, it follows that $\angle MJI=90^{\circ }$.

In the triangles $GIK$ and $HMN$, line segments $[GJ]$ and $[HJ]$ are angle bisectors and altitudes, therefore they are also medians. Thus, the diagonals of the quadrilateral $MKNI$ are perpendicular and bisect each other.

b) The sides of the rhombus are parallel to $AC$ and $BD$, respectively. Indeed, according to the angle bisector theorem, $\frac{MB}{MC}=\frac{GB}{GC}$ and $\frac{BK}{KA}=\frac{HB}{HA}$. But from the power of the point $G$ with respect to the circle, $\frac{GB}{GC}=\frac{GD}{GA}$. In order to prove that $MK\parallel AC$, i.e. $\frac{BK}{KA}=\frac{BM}{MC}$, it is sufficient to prove that $HA\cdot GD=HB\cdot GA$. This follows from $[AHG]=\frac{HA\cdot GD\sin D}{2}=\frac{HB\cdot GA\sin B}{2}$ and $\sin B=\sin D$. Thus, $MK\parallel AC$.

Consider $\{L\}=MK\cap BF$, $\{P\}=AE\cap KN$, $\{O\}=DF\cap IN$ and $\{Q\}=MI\cap CE$. Line segments $AE,BF,CE$ and $DF$ are medians in triangles $ABD,ABC,BCD,CDA$, therefore $LO$ and $PQ$ meet in the center of the rhombus.

Put $\{X\}=LO\cap EF$ and $\{Y\}=QP\cap EF$. Then $\frac{XE}{XF}=\frac{LB}{LF}=\frac{BK}{KA}$ and $\frac{YE}{YF}=\frac{QE}{QC}=\frac{BM}{MC}$. As $\frac{BK}{KA}=\frac{BM}{MC}$, it follows that points $X$ and $Y$ coincide, which means that the lines $EF,LO$ and $PQ$ pass through the center of the rhombus.