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.