Korean Mathematical Olympiad Final Round

Let $O_1$, $O_2$ be the circumcenters of the triangles $AED$ and $EBC$, respectively.

Lemma Let $H$ be the foot of the altitude of the triangle $ABC$. The line symmetric to the line $AH$ with respect to the bisector of the angle $A$ passes the circumcenter $O$ of the triangle $ABC$.

Proof Since the triangle $BOA$ is isosceles and $\angle BOA=2\angle C$, we have $$\angle BAO=\frac{180^{\circ }-2\angle C}{2}=90^{\circ }-\angle C.$$ And since $\angle CAH=90^{\circ }-\angle C$, we have $\angle CAH=\angle BAO$ and thus the lines $AH$ and $AO$ are symmetric with respect to the angle bisector of the angle $A$. $\diamond$

We have $\angle BEP=90^{\circ }-\angle PBE=90^{\circ }-\angle EDA=\angle DER$. So by the above lemma $O_1$ lies on the line $EQ$ and similarly $O_2$ lies on the line $ES$.

Since $△EAD\sim △ECB$ we have $$E{O}_1:EQ=E{O}_2:ES.$$ So we have $△E{O}_1{O}_2\sim △EQS$, and thus $O_1{O}_2\parallel QS$. Therefore the three points $E$, $K$, $M$ lie on the same line, where $M$ is the midpoint of $O_1{O}_2$.

Now we are going to show that the three points $E$, $M$, $O$ lie on the same line. Since $O_1$ is the circumcenter of the $△AED$ and $O$ is the center of the circun circle of the quadrilateral $ABCD$, the two points $O$, $O_1$ lie on the perpendicular bisector of $AD$. So $O{O}_1\perp AD$ and $E{O}_2\perp AD$ and thus we have $O_1{O}_2\parallel E{O}_2$. Similarly, since $O_2$ is the circumcenter of the $△EBC$ and $O$ is the center of circun circle of the quadrilateral $ABCD$, the two points $O$, $O_2$ lie on the perpendicular bisector of $BC$. So $E{O}_1\perp BC$ and $E{O}_1\perp BC$ and thus we have $E{O}_1\parallel O{O}_2$. So the quadrilateral $O{O}_{1}E{O}_2$ is a parallelogram and the line $OE$ passes the midpoint $M$ of $O_1{O}_2$, and thus three points $E$, $M$, $O$ lie on the same line. Therefore three points $E$, $K$, $O$ lie on the same line. $\square$