[asy][asy]size(7.5cm); pair A,B,O,K,L,M,P,C,X; path k1=circle((0,0),10); O=(10,0); path k2=circle(O,6.5); pair[] x=intersectionpoints(k1,k2); A=x[0]; B=x[1]; K=O*dir(75); L=intersectionpoint((O+(K-O)*0.99)--O,k2); M=2*O-L; P=foot(L,A,B); C=(A+B)/2; X=extension(K,O,A,B); draw(k1^^k2,green); filldraw(K--L--P--cycle,lightmagenta); filldraw(C--O--M--cycle,lightmagenta); D(MP("K",K,N)--MP("L",L,WNW)--MP("X",X,E)--MP("O",O,E)--MP("M",M,SE)--MP("B",B,S)--MP("C",C,W)--MP("P",P,E)--MP("A",A,N)); label("$k_1$",(0,0)-B,NW);label("$k_2$",2*O-B,NE); dot(O);dot(A);dot(K);dot(L);dot(M);dot(P);dot(C);dot(X);dot(B); [/asy][/asy] Let $C$ be the midpoint of $AB$ and let $X=KO\cap AB$. Obviously, $CO\perp AB$, We'll first prove that $\triangle KLP\stackrel{+}{\sim}\triangle MOC$. Since $LP||CO$, and $KL||MO$ (!), we already have $\angle KLP=\angle MOC$. It suffices to prove $\tfrac{KL}{MO}=\tfrac{LP}{OC}$. But from similar triangles, $\tfrac{LP}{OC}=\tfrac{LX}{OX}$. So we need $$\frac{KL}{MO}=\frac{LX}{OX}\iff \frac{KL}{OL}=\frac{LX}{OX}\iff \frac{OL}{OX}+1=\frac{KL}{LX}+1\iff \frac{OL+OX}{OX}=\frac{KX}{LX}.$$ But since $X$ lies on the radical axis of the two circles, we have from power-of-point: $$KX\cdot OX=OL^2-OX^2=(OL+OX)(OL-OX)=(OL+OX)(LX)\implies \frac{OL+OX}{OX}=\frac{KX}{LX}.$$Therefore $\triangle KLP\stackrel{+}{\sim}\triangle MOC\implies \angle LKP=\angle OMC\implies KP||MC$, as desired. $\blacksquare$

Well I have a simpler one - Let AB and KM meet at N. Let Q be the mid point of AB. Join OQ. It is easy to see that OQ is parallel to PL. So, QN/PN=ON/LN ........1 By power of a point theorem, MN•LN=AN•BN=KN•ON or MN•LN=KN•ON or MN/KN=ON/LN or MN/KN=QN/PN ( from 1) This implies that triangles MNQ and KNP are similar. So, KP is parallel to QM.

Ankoganit wrote: Circles $k_1$ and $k_2$ intersect in points $A$ and $B$, such that $k_1$ passes through the center $O$ of the circle $k_2$. The line $p$ intersects $k_1$ in points $K$ and $O$ and $k_2$ in points $L$ and $M$, such that the point $L$ is between $K$ and $O$. The point $P$ is orthogonal projection of the point $L$ to the line $AB$. Prove that the line $KP$ is parallel to the $M-$median of the triangle $ABM$. Matko Ljulj Claim : $L$ is the incentre of $\triangle AKB$. Proof : Note that $OA = OB$. So, $KL$ bisects $\measuredangle{KAB}$. Also, note that $\measuredangle{ALB}$ = $180^{\circ}$ - $\measuredangle{AMB}$ = $180^{\circ}$ - $\frac{ \measuredangle{AOB}}{2}$ = $90^{\circ}$ + $\frac{\measuredangle{AKB}}{2}$. So, $ L$ is the incentre of $\triangle AKB$. $\square$ Now it isn’t hard to see that $M$ is the $K - \text{excentre}$ of $\triangle KAB$. Now the desired claim follows from a well known result. $\blacksquare$

EMC 2015 Seniors P3 wrote: Circles $k_1$ and $k_2$ intersect in points $A$ and $B$, such that $k_1$ passes through the center $O$ of the circle $k_2$. The line $p$ intersects $k_1$ in points $K$ and $O$ and $k_2$ in points $L$ and $M$, such that the point $L$ is between $K$ and $O$. The point $P$ is orthogonal projection of the point $L$ to the line $AB$. Prove that the line $KP$ is parallel to the $M-$median of the triangle $ABM$. Solution: By Fact $5$, $L$ is the incenter of $\Delta KAB$. Let $KM \cap AB=R$ and Let $M'$ be midpoint of $AB$ $\implies$ $LP || M'O$. Therefore, we have, $\Delta LRP$ $\sim$ $\Delta ORM'$. Now, Observe that $K$ is the image of $R$ under $(O)$ $$\therefore ~ \qquad \frac{RP}{RM'}=\frac{LR}{OR}=\frac{KR}{RM} \implies KP||MM' \qquad \blacksquare$$

ok i have a complex bash solution since nobody posted this i am going to post it \[\color{blue} a=a,b=\overline{a}=1/a,n=(a^2+1)/2a,l=-m\] we consider that $KLA~DNM$ then we prove that $KAOB$ is cyclic. let's compute $k$: $k=\frac{m(p-l)}{n}+l=(-a^4 m+ a^2 m + 2 a + 2 m)/(a^4 + a^2)$ so from the cyclicity condition we should prove that \[\frac{k+a}{k}=\frac{\overline{k}+\overline{a}}{\overline{k}}\]so we should see that: \[\frac{(2 ((a^2 + 1) m + a)-m((a^2 (a^2 + 1)) ))/(a^2 (a^2 + 1)) +a}{(2 ((a^2 + 1) m + a)-m((a^2 (a^2 + 1)) ))/(a^2 (a^2 + 1)) }=\frac{(-(1/a)^4 /m+ (1/a)^2 /m + 2 (1/a) + 2 /m)/((1/a)^4 + (1/a)^2)+1/a}{(-(1/a)^4 /m+ (1/a)^2 /m + 2 (1/a) + 2 /m)/((1/a)^4 + (1/a)^2)}\]which is easy to check. \[\color{magenta} \boxed{\mathbf{Q.E.D}}\]

The following more general and symmetric claim is true: Let $k_1$ and $k_2$ be two circles that intersect at points $A$ and $B$ (but $k_1$ does not have to pass through the center of $k_2$ as in the given problem). A line $\ell$ intersects $k_1$ in points $M$ and $G$ and $k_2$ in points $N$ and $K$ such that the four points line in the order $M$, $N$, $G$, $K$ on $\ell$. Let $P$ be the foot of the perpendicular $G$ to $AB$, and let $J$ be the foot of the perpendicular from $N$ to $AB$. Then $KP \parallel MJ$. Proof: Let $\ell$ intersect $AB$ at $E$. It is easy to see that $\triangle NEJ \sim \triangle GEP$, so $\frac{NE}{EJ} = \frac{GE}{EP}$. Next, note that $NE \cdot KE = AE \cdot EB = ME \cdot GE$. Combining the two, $\frac{ME}{EJ} = \frac{KE}{EP}$ Thus, $\triangle MEJ \sim \triangle KEP,$ so $KP \parallel MJ$.