Saudi Arabian Mathematical Competitions

First, we know that $EI = EB = EC$ then $E$ is the circumcenter of triangle $BIC$. Thus $KI$ is the tangent line of circle $(BIC)$. Hence,
$$KI^2 = KB \cdot KC = KE \cdot KN.$$
By combining with $KIE$ is right triangle, we get $IN \perp KE$.

So $PE$ is the diameter of circle $(O)$. We have $\triangle BDE \sim \triangle ABE$ so
$$\frac{BE}{AE} = \frac{DE}{BE} \Rightarrow EA \cdot ED = EB^2 = EI^2 = EN \cdot EK$$
Implies that $ADNK$ is cyclic. Thus $\angle MAE = \angle DNE$ which leads to $EM, EQ$ are equal. Therefore, $MPQE$ is a kite.

Since $KI^2 = KM \cdot KA$, we get $IM \perp AK$. Construct the diameter $AA'$ of $(O)$ then $M, I, A'$ are collinear. Thus $AJ \parallel IM$ implies that $AIA'J$ is a parallelogram and then $O$ is the midpoint of $IJ$.

Since $IO$ is the median of triangle $IPE$ so $Ix, IO, IP, IE$ form a harmonic quartet with $Ix \parallel PE$. Denote $Ky$ as the ray that perpendicular to $IO$ then notice that $KE \perp IP, KI \perp IE, KD \perp Ix, K, Ky \perp IO$ so $KD, Ky, KI, KE$ form another harmonic quartet. Since $Ky \parallel RS$ then $R$ is the midpoint of $OS$. Therefore $OR$ is the midline of triangle $IJS$ which implies that $OR \parallel JS$.
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