Let $P,Q,R,S$ be the midpoints $AB,BC,CD,DA$ and $U$ the orthogonal projection of $K$ on segment $CD.$ Then we have $\angle UKD=\angle KCD=\angle KBP=\angle PKB$ $\Longrightarrow$ $P,K,U$ are collinear, i.e. $PK \perp CD.$ Likewise, we'll have $RK \perp AB.$ Since $OR \perp CD$ and $OP \perp AB,$ it follows that $PORK$ is a parallelogram $\Longrightarrow$ $PR$ passes through the midpoint of $OK.$ Similarly, $SQ$ passes through the midpoint of $OK.$ Hence, $G$ is the midpoint of segment $OK.$ Now, $BD$ is the radical axis of $(O), \odot(OBD)$ and $AC$ is the radical axis of $(O), \odot(OAC)$ $\Longrightarrow$ $OGK$ is the radical axis of $\odot(OBD),$ $ \odot(OAC).$ Thus, 2nd intersection $L$ of $\odot(OBD)$ and $\odot(OAC)$ lies on the line $OGK.$

A nice generalization of the "hard" part of the problem: Quote: Let $ABCD$ be a quadrilateral whose diagonals $AC$ and $BD$ are perpendicular. Let $O$ be the intersection of the perpendicular bisectors of $AC$ and $BD$ and let $G$ be the intersection of the diagonals of the quadrilaterals whose vertices are the midpoints of the sides of $ABCD$. Prove that $G$ is the midpoint of $OK$.

Joao Pedro Santos wrote: Let $ABCD$ be a quadrilateral whose diagonals $AC$ and $BD$ are perpendicular. Let $O$ be the intersection of the perpendicular bisectors of $AC$ and $BD$ and let $G$ be the intersection of the diagonals of the quadrilaterals whose vertices are the midpoints of the sides of $ABCD$. Prove that $G$ is the midpoint of $OK$. I solved the "easy" part of the problem (ie, $K\in OL$) saying that $OL$ is clearly the radical axis of the circumcirles of $OAC$ and $OBD$, and $K$ is on that radical axis since $KA\cdot KC=KB\cdot KD$ because $ABCD$ is cyclic. Note that we do not need the diagonals of $ABCD$ to be perpendicular for this part, but if $K$ is on the radical axis, then necessarily $KA\cdot KC=KB\cdot KD$, ie $K\in OL$ iff $ABCD$ is cyclic, regardless of the angle between the diagonals. I propose a different proof (no coordinates) for the "improved" version of the "hard" part of the problem (ie $G$ is the midpoint of $OK$). Denote $M$ the midpoint of $OK$, and $T$ the midpoint of $AB$. Clearly $KT=TA=TB=\frac{AB}{2}$ is the circumradius of right-angled triangle $AKB$, and denote by $r$ the circumradius of $ABCD$. Applying the median theorem twice, \[MT^2=\frac{OT^2+KT^2}{2}-\frac{OK^2}{4}=\frac{OA^2+OB^2}{4}-\frac{AB^2}{8}+\frac{AB^2}{8} -\frac{OK^2}{4}=\frac{r^2}{2}-\frac{OK^2}{4}.\] By cyclic symmetry, the result is the same for the midpoints $U,V,W$ of $BC,CD,DA$, or $MT=MU=MV=MW$. But $GT=GU=GV=GW$, since the midpoints of the sides of $ABCD$ form a parallelogram whose sides are parallel to diagonals $AC,BD$, which are in turn perpendicular, ie, $TUVW$ is a rectangle and its diagonals meet at point $G$ which is at the same distance from all vertices.

Since $ABCD$ is cyclic $AK\cdot KC=BK\cdot KD$ and since $K$ is inside the circumcircles of $AOC$ and $BOD$ we get that the power of $K$ with respect to them is the same, this implies the collinearity of $O$, $K$ and $L$. Now it's really easy to prove that the perpendicularity of the diagonals of $ABCD$ implies that its anticenter is $K$. Now, the center, barycenter and anticenter of a cyclic quadrilateral are collinear. Done.

Joao Pedro Santos wrote: A nice generalization of the "hard" part of the problem: Let $ABCD$ be a quadrilateral whose diagonals $AC$ and $BD$ are perpendicular. Let $O$ be the intersection of the perpendicular bisectors of $AC$ and $BD$ and let $G$ be the intersection of the diagonals of the quadrilaterals whose vertices are the midpoints of the sides of $ABCD$. Prove that $G$ is the midpoint of $OK$. An elementary proof: let $P, Q, R, S$ be the midpoints of $AB, BC, CD, DA$. Note that $PQ$ and $RS$ are midsegments in $\bigtriangleup ABC$ and $\bigtriangleup ACD$, both opposing $AC$, thus $PQ\parallel AC\parallel RS$ and similarly $QR\parallel BD\parallel SP$. Since $AC\perp BD$, both pairs of sides in $\Box PQRS$ are perpendicular, meaning it's a rectangle. Now let $K_A$ be the intersection of $KA$ with $SP$, $K_B$ the intersection of $KB$ with $PQ$, $K_C$ the intersection of $KC$ with $QR$, $K_D$ the intersection of $KD$ with $RS$, and $K'$ the reflection of $K$ with respect to the center $G$ of the rectangle. It's easy to see that $K_A$ is the midpoint of $KA$, given that the midsegment $SP$ of $\bigtriangleup ABD$ cuts the altitude $AK$ on $K_A$. Similarly $K_C$ is the midpoint of $KC$. Now note that $G$ is the midpoint of $KK'$. Then $GK_A$ is the midsegment opposing $AK'$ in $\bigtriangleup KAK'$, so $AK'=2GK_A$. This way we also have $CK'=2GK_C$. Also, since $K_AK_C$ is an altitude parallel to $PQ$ in rectangle $\Box PQRS$, they share their perpendicular bisector, so $G$ must lie in the perpendicular bisector of $K_AK_C$ and $\bigtriangleup GK_AK_C$ must be isoceles on $G$. It follows that $GK_A=GK_C$ and $AK'=CK'$, $\bigtriangleup ACK'$ is isoceles on $K'$, and $K'$ lies on the perpendicular bisector of $AC$. It also follows that $K'$ lies in the perpendicular bisector of $BD$. Since $AC$ and $BD$ are perpendicular, so are their perpendicular bisectors, meaning $K'$ is their unique intersection. Yet $O$ is the circumcircle of $\bigtriangleup ABD$ and $\bigtriangleup ABC$, so it must lie on both the perpendicular bisectors of $AC$ and $BD$, and thus must coincide with their intersection. Hence $K'=O$ and we are done.

Well for the first part , Power of a point\radical axis does the job. Then for the second part ,we use complex numbers. Finding $G$ is a bit work but direct computation. Then we just prove $G-K-L$ collinear ,also using $bd+ac=0$.

Let $W, X, Y, Z$ be the midpoints of $\overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}$ respectively. Since $\overline{WZ}\parallel\overline{XY}\parallel\overline{BD}\perp\overline{AC}\parallel\overline{WX}\parallel\overline{YZ}$, we know $WXYZ$ is a rectangle. Thus, $G$ is its center. Let $h$ be the homothety centered at $G$ with ratio $-1$. Then, $h(W)=Y, h(X)=Z$. Let $h(B)=B'$. Note that $\angle YB'Z = \angle WBX$. Furthermore, \[ \angle YKZ=\angle YKD+\angle DKZ = \angle YDK + \angle KDZ = \angle ADC. \]This means that $\angle YB'Z + \angle YKZ = \angle ABC +\angle ADC = 180^{\circ}$. Thus, $K\in (YB'Z)$. Also, note that $O\in (WBX)$, since $\angle OWB=\angle OXB=90^{\circ}$. Furthermore, $h((WBX))=(YB'Z)$. Doing the same for the other vertices we can conclude $h(O)=K$. This tells us $K, G, O$ are collinear. Now, note that by PoP, $KA\cdot KC=KB\cdot KD$. Thus, $K$ has the same power to both $(OAC)$ and $(OBD)$, so it is on their radical axis, which is $\overleftrightarrow{OL}$. Thus, the four desired points are collinear. $\blacksquare$