Here is what I have so far: I will first prove the converse. We will assume KBCD is cyclic. Let KC and BD intersect at P. Then ABCP is also cyclic. By power of a point, $ MA \cdot MC = MP \cdot MB$. So we need to show $ MA \cdot CD = MB(MD - MP) = MB \cdot PD$. Or $ \frac {MA}{MB} = \frac {PD}{CD} \implies \frac {MP}{PC} = \frac {PK}{KB}$. But by Power of a point, $ PC \cdot PK = PD \cdot PB$. So we need to show $ \frac {KB}{PB} = \frac {PD}{PM} \implies \frac {CD}{CP} = \frac {CD}{CM}$ (angle bisector theorem). But then that would mean CP=CM, where's my mistake?

dgreenb801 wrote: Or $ \frac {MA}{MB} = \frac {PD}{CD} \implies \frac {MP}{PC} = \frac {PK}{KB}$. Here, I think. And if you prove the converse, you have to show it implies the original problem.

Note:$ MH.CD=HD.MC$(1) $ MH+HD=MD \longrightarrow MC.MH+MC.HD=MC.MD \longrightarrow$ according to(1),$ MC.MH+CD.MH=MC.MD$ multiple the equation at $ \frac{MA}{MH}$,then,$ MA.MC+MA.CD=\frac{MA.MC.MD}{MH}$(2) according to assumption we have: $ \frac{MA.MC.MD}{MH}=MB.MD \longrightarrow MA.MC=MB.MH$ $ ABCH$ is cyclic,then $ \angle ACH=\angle ABH=\angle HCD \longrightarrow KBCD$ is cyclic and $ \angle BKC=\angle CDB$

(1) draw a circle with center $ C$ and radius $ CD$ (2) let the intersection of that circle and $ AC$ be $ E$ (3) $ MA\cdot MC$+$ MA\cdot CD$=$ AM\cdot ME$=$ BM\cdot MD$ $ \Longrightarrow$ $ A,B,E,D$ are concyclic (4) $ \angle ABD$=$ \angle KCD$ $ \Longrightarrow$ $ K,B,C,D$ are concyclic (5) $ \Longrightarrow\angle BKC$=$ \angle CDB$

77ant wrote: (1) draw a circle with center $ C$ and radius $ CD$ (2) let the intersection of that circle and $ AC$ be $ E$ (3) $ AM\cdot MC$+$ AC\cdot CD$=$ AM\cdot ME$=$ BM\cdot MD$ $ \Longrightarrow$ $ A,B,E,D$ are concyclic (4) $ \angle ABD$=$ \angle KCD$ $ \Longrightarrow$ $ K,B,C,D$ are concyclic (5) $ \Longrightarrow\angle BKC$=$ \angle CDB$ How did you get (3) $ AM\cdot MC$+$ AC\cdot CD$=$ AM\cdot ME$. AM*MC+AC*CD=AM*ME=AM(MC+CE)=AM*MC+AM*CE=AM*MC+AM*CD ----> AC=AM which I think is impossible.

My solution(77ant`s idea): Draw a circle with center C and radius CD. Let the intersection of that circle and AC be E(E lies on the extension of AC). MA*MC+MA*CD=MA(MC+CD)=MA(MC+CE)=MA*ME MA*ME=MB*MD so A,B,E,D are concyclic. <CDE=<CED=<ACD/2=x <AED=<ABD=<KCD=x, so K,B,C,D are concyclic. ---> <BKC=<CDB.

my mistake in typing I have corrected it right now.

Very interesting question Note that showing <BKC = <CDB is the same as showing KBCD cyclic. Extend AC and let a point D' be on AC with CD = CD'. Now the metric condition given becomes MA(MC+CD') = MB(MD) so MA(MD') = MB(MD). This implies that quadrilateral ABD'D is cyclic. Now let <CDD' = <CD'D = <AD'D = <ABD (by cyclic quads) = <1. Since <DCA is an exterior angle, it is the sum of <CDD' and <CD'D, so <DCA = 2<1. Since CK is the angle bisector of <DCA, <DCK = <KCA = <1. Now let X be the intersection of CK and BD. Since <XCA = <XBA, quadrilateral BAXC is cyclic as well. Now this question is just a simple angle chase. Let <CAB = <CXB = <2. Then <CKA = <CAB - <KCA (by exterior angles) = <2-<1. <CDM = <CXM - <DCX (by exterior angles again) = <2-<1. Thus, <CKA = <CDM, so quadrilateral KBCD is cyclic. Finally, this implies that <BKC = <CDB. QED (Sorry for no diagram or LaTeX, I don't know how to do either )

Solution: Let $BD$ meet $KC$ at $N$. $ MA \cdot MC +MA \cdot CD = MB \cdot MD$ implies that $\frac{MB}{MA}$ = $\frac{MC+CD}{MD}$. Applying angle bisector theorem, we get $\frac{MC}{MN}$ = $\frac{CD}{DN}$, or, $\frac{MC\cdot ND}{MN\cdot MD}$ = $\frac{CD}{MD}$, or, $\frac{MC}{MN}$ = $\frac{MC+CD}{MD}$. Thus, $\frac{MC}{MN}$ = $\frac{MB}{MA}$, or, $MC\cdot MA$ = $MB\cdot MN$. Hence, by the converse of angle bisector theorem, $ABCN$ is cyclic. This implies that $\angle ABD$ = $\angle ACN$, or, $\angle KBD$ = $\angle KCD$, or, $KBCD$ is cyclic. So, $\angle BKC$ = $\angle CDB$.

Nice one. Let $\omega$ be the circumcircle of $BCD$. Let $E=AC\cap \omega$ and $K'$ be the point on $\omega$, that lies on angle bisector of $\angle ACD$. We ultimately want to show that $A,B,K'$ are collinear. We have that $$ MA \cdot MC +MA \cdot CD = MB \cdot MD, \quad (1)$$but by Power of a Point we have that $$MB\cdot MD=MC\cdot ME.$$Thus, $(1)$ takes the following form: $$MA \cdot MC +MA \cdot CD =MC\cdot ME \Longleftrightarrow MA\cdot CD=AE\cdot MC\Longleftrightarrow \frac {MA}{AE} = \frac {MC}{CD} \quad (2)$$We also have that $\triangle MCD\sim \triangle MBE$, since $\angle ECD=\angle EBD$ and $\angle EMB=\angle DMC$, thus $$\frac {MC}{CD} = \frac {MB}{BE} \quad (3)$$By $(2)$ and $(3)$, we conclude that $\frac {MA}{AE} = \frac {MB}{BE}$, which means that $AB$ is the angle bisector of $\angle DBE$. Also, since $K'$ is the defined as intersection of $\omega$ and angle bisector of $\angle ACD$, we have that $\triangle K'DE$ is isosceles, from here we have that $K'B$ is the angle bisector of $\angle DBE$. We conclude that $A,B,K'$ are collinear.

since it came back in front page due to the last solution, let's say the origin of the problem: 2000 Belarusian MO B 10.3 also posted and solved here, here and here

Let $D'$ be on $AC$ such that $C$ lies between $M,D'$ and $CD' = CD$. $MA.MC + MA.CD = MB.MD \implies MA.MD' = MB.MD \implies ABD'D$ is cyclic. $\angle KBD = \angle ABD = \angle AD'D = \frac{\angle ACD}{2} = \angle KCD \implies KBCD$ is cyclic so $\angle BKC = \angle BDC$.

Let $N= BD\cap KC$. By the angle bisector theorem, we know that $\frac{NM}{ND}= \frac{MC}{CD}$.(1) As $MA(MC+CD)= MB\cdot MD$, by (1), we get $MA(MC+\frac{MC\cdot ND}{NM})= MB\cdot MD$ $\implies MA\cdot MC(1+\frac{ND}{NM})= MB\cdot MD$ $\implies MA\cdot MC\cdot\frac{(ND+NM)}{NM}= MB\cdot MD\implies MA\cdot MC\cdot\frac{MD}{NM}= MB\cdot MD$ $\implies MA\cdot MC= NM\cdot MB$. So, $ABCN$ is ciclic $\implies \angle DCN=\angle NCM= \angle NBA\implies KBCD$ is ciclic, therefore $\angle BKC=\angle BDC$, QED

Let the angle bisector of $\angle{ACD}$ intersect $BD$ at $X$. By given condition, we know that $MA=\frac{MB \cdot MD}{MC+CD}$. By angle bisector theorem, we get that $XM=\frac{MC \cdot MD}{MC+CD}$. Then $XM \cdot MB=\frac{MB \cdot MC \cdot MD}{MC+CD}=AM \cdot MC$, implying that $ABCX$ is cyclic by converse of PoP. Therefore $\angle{DCX}=\angle{MCX}=\angle{ABD}$, implying that $BCDK$ is cyclic, and $\angle{BKC}=\angle{BDC}$ $\blacksquare$

Extend $AC$ to meet $(ABD)$ at $D'$. Then $MD = MD'$, and from here some angle chasing shows us that $BKDC$ is cyclic as desired.

Cute and easy!! Denote $\boxed{MA=x,MB=w,MC=y,CD=z}$. Then by the given condition, we have $xy+xz=w(MD) \implies \boxed{MD=\frac{x(y+z)}{w}}$. Define $O=BD \cap KC$ then $CO$ is the angle bisector of $\angle{MCD}$. By internal angle bisector theorem, we have $\frac{MO}{OD}=\frac{MC}{CD}=\frac{y}{z}$ and $MO+OD=MD=\frac{x(y+z)}{w}$. Thus, $\boxed{MO=\frac{xy}{w}}$ and $\boxed{OD=\frac{xz}{w}}$. Now note that $MO.MB=xy=MA.MC$ implying $BAOC$ is cyclic. Thus, $\angle{KBD}=\angle{ABO}=\angle{ACO}=\angle{ACK}=\angle{KCD} \implies \boxed{\angle{KBD}=\angle{KCD}}$. Thus $BKDC$ is also cyclic implying $\angle{BKC}=\angle{CDB}$ as desired. $\blacksquare$ ($\mathcal{QED}$)