VMO

a) It is easy to see that $K$ belongs to segment $AC$. Since $IK=IC$, we have triangle $IKC$ is isosceles at $I$. Hence, $\angle AKI=180^{\circ }-\angle IKC=180^{\circ }-\angle ICK=\angle ABI$, because $ABIC$ is a cyclic quadrilateral. Furthermore, note that $I$ is the midpoint of arc $BC$, then $\angle IAK=\angle IAB$ and $IK=IB=IC$ so $△ABI=△AKI$.



We conclude that $AI$ is the perpendicular bisector of the segment $BK$, which implies that $E$ is the midpoint of $BK$. Note that, $$\angle DCK=\angle ABD=\angle AKB=\angle DKC$$ so triangle $DKC$ is isosceles, hence $DK=DC$. Note that $IK=IC$, we can see that $ID$ is also the perpendicular bisector of $KC$ so $F$ is the midpoint of $CK$. From these above results, we have $EF$ is the midline of triangle $KBC$, then $EF=\frac{1}{2}BC$.

b) Let $J$ be the midpoint of arc $BC$ of $(O)$ that contains $A$, it is well-known that $IJ$ is the diameter of circle $(O)$. We will show that $J$, $K$ and $P$ are collinear.

Indeed, from $\angle IPJ=90^{\circ }$, we only need to prove $\angle KPI=90^{\circ }$. In triangle $ADI$, because $DK\perp AI$, $AK\perp DI$ then $K$ is the orthocenter of triangle $ADI$. Hence, $IK$ is perpendicular to $AD$.

On the other hand, note that $CM\parallel AD$, it follows that $CM\perp IK$, but $IM\perp KC$ so $M$ is the orthocenter of triangle $IKC$. Therefore, $$\angle MKC=90^{\circ }-\angle KCI=90^{\circ }-(\angle ACB+\frac{1}{2}\angle BAC),\text{ so}$$ $$\angle KNB=\angle NKC+\angle NCK=90^{\circ }-\frac{1}{2}\angle BAC.$$ We have $\angle KPI=\angle KPB+\angle BPI=\angle KNB+\angle IAB=90^{\circ }-\frac{1}{2}\angle BAC+\frac{1}{2}\angle BAC=90^{\circ }$. Hence, $J$, $P$ and $K$ are collinear.

We also have $AJ\perp AI$, $KD\perp AI$ so $AJ\parallel KD$ and $DJ\perp ID$, $AK\perp ID$ then $DJ\parallel AK$. Therefore, $AJDK$ is a parallelogram and the line $PK$ bisects the segment $AD$.