Let $M$ be the midpoint of $AK$. Suppose the tangents to $\omega$ at $A$ and $K$ intersect at $T$. [asy][asy] size(6cm); defaultpen(fontsize(9pt)); import geometry; pair A = dir(130), B = dir(210), C = dir(-30), N = dir(-90), K = extension(A, N, B, C), M = (A+K)/2, O = extension(K, rotate(90, K)*C, A, (0, 0)), T = extension(O, M, B, C); draw(unitcircle, blue); draw(circumcircle(B, O, C), dashed+heavygreen); draw(circle(O, abs(O-K)), heavycyan); draw(T--A, fuchsia); draw(T--O, magenta); draw(T--C, fuchsia); draw(A--N, orange); string[] names = {"$A$", "$B$", "$C$", "$M$", "$N$", "$O$", "$K$", "$T$"}; pair[] points = {A, B, C, M, N, O, K, T}; pair[] ll = {A, dir(225), C, dir(160), N, dir(-40), dir(240), T}; for (int i=0; i<names.length; ++i){dot(names[i], points[i], dir(ll[i]));} [/asy][/asy] Since $\overline{TA}$ and $\overline{TK}$ are tangents to $\omega$, $TA = TK$ and $\overline{TM} \perp \overline{AK}$. Therefore, $T,M,O$ are collinear. By power of a point and $\triangle TAO \sim \triangle TMA$, we have \[ TB \cdot TC = TA^2 = TM \cdot TO, \]implying that $BMOC$ is cyclic.

[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.46289922013448, xmax = 17.741598611080327, ymin = -15.083920169946229, ymax = 16.51990056839753; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); /* draw figures */draw(circle((1.3928326504947515,2.022170021992736), 4.572369873409059), linewidth(0.8) + qqwuqq); draw(circle((-0.2109752479022542,-0.765101142599848), 7.788124888942724), linewidth(0.8) + qqwuqq); draw(circle((-5.203432588776878,-3.0773971741628325), 8.337643593661772), linewidth(0.8) + linetype("4 4") + blue); draw((-8.807983617951303,13.167038683977141)--(1.1928456724961864,-8.425660920398126), linewidth(0.8) + fuqqzz); draw((-8.807983617951303,13.167038683977141)--(1.3928326504947515,2.022170021992736), linewidth(0.8) + fuqqzz); draw((-8.807983617951303,13.167038683977141)--(3.6732311516066267,5.98529365561628), linewidth(0.8) + fuqqzz); draw((-2.7561333029429687,0.10054368566368699)--(3.6732311516066267,5.98529365561628), linewidth(0.8)); /* dots and labels */dot((1.3928326504947515,2.022170021992736),dotstyle); label("$O$", (1.8730135146123619,1.6847737450188671), NE * labelscalefactor); dot((3.6732311516066267,5.98529365561628),dotstyle); label("$A$", (3.939930060790962,6.518691473984949), NE * labelscalefactor); dot((-2.7561333029429687,0.10054368566368699),dotstyle); label("$K$", (-3.6943262146106424,-0.4154801648077751), NE * labelscalefactor); dot((1.1928456724961864,-8.425660920398126),linewidth(4pt) + dotstyle); label("$B$", (0.7395431505789358,-9.4165683497791), NE * labelscalefactor); dot((-5.145798095971687,5.260047216066601),linewidth(4pt) + dotstyle); label("$C$", (-6.427953760917704,5.451895837247607), NE * labelscalefactor); dot((-8.807983617951303,13.167038683977141),linewidth(4pt) + dotstyle); label("$E$", (-8.528243943576724,13.886248840202217), NE * labelscalefactor); dot((0.45854892433182903,3.0429186706399833),linewidth(4pt) + dotstyle); label("$M$", (-0.8606503045270774,2.8182441090522934), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let the tangent at $A$ to both circles meet $BC$ at $E$. Let $M$ be the midpoint of $AK$. Then we have $E,M,O$ are collinear since $E$ is the intersection of tangents. Moreover we have $\triangle EAM\sim\triangle EOA$ Therefore, $$EM\times EO=EA^2=EB\times EC$$by power of a point theorem we have $(BOC)$ passes through $M$, so the circle $(BOC)$ bisects the segment $AK$ as desired.

Just invert wrt $\omega$ and apply the theorem about radical center of three circles.

NJOY wrote: Let circles $\Omega$ and $\omega$ touch internally at point $A$. A chord $BC$ of $\Omega$ touches $\omega$ at point $K$. Let $O$ be the center of $\omega$. Prove that the circle $BOC$ bisects segment $AK$. Kinda easy for Problem 16? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -27.723877902894355, xmax = 25.42130926711064, ymin = -17.645498863838338, ymax = 12.998348763021179; /* image dimensions */ pen qqwwxq = rgb(0,0.4,0.4392156862745098); pen ttqqtt = rgb(0.2,0,0.2); pen qquvtu = rgb(0,0.27058823529411763,0.20392156862745098); pen avqsry = rgb(0.6470588235294118,0.00784313725490196,0.09411764705882353); pen zzwwqq = rgb(0.6,0.4,0); pen qqzzqq = rgb(0,0.6,0); pen qqwwtt = rgb(0,0.4,0.2); draw((-5.253612464770877,7.831239053966863)--(-6.417914253354641,-2.6125144798721553)--(6.902085746645359,-2.432514479872155)--cycle, linewidth(0.4) + qqwwxq); /* draw figures */ draw((-5.253612464770877,7.831239053966863)--(-6.417914253354641,-2.6125144798721553), linewidth(0.4) + qqwwxq); draw((-6.417914253354641,-2.6125144798721553)--(6.902085746645359,-2.432514479872155), linewidth(0.4) + qqwwxq); draw((6.902085746645359,-2.432514479872155)--(-5.253612464770877,7.831239053966863), linewidth(0.4) + qqwwxq); draw(circle((0.1818016913243761,1.9385056138805896), 8.01673464973063), linewidth(0.4) + ttqqtt); draw(circle((-1.2002582691666517,3.4368479133826457), 5.9783236556301835), linewidth(0.4) + qquvtu); draw((-5.253612464770877,7.831239053966863)--(-16.72710660684508,-2.7518278900544586), linewidth(0.4) + avqsry); draw((-16.72710660684508,-2.7518278900544586)--(-6.417914253354641,-2.6125144798721553), linewidth(0.4) + zzwwqq); draw((-1.1332933334831907,-2.5411006836576764)--(-5.253612464770877,7.831239053966863), linewidth(0.4) + qqzzqq); draw((-1.2002582691666517,3.4368479133826457)--(-16.72710660684508,-2.7518278900544586), linewidth(0.4) + qqwwtt); draw(circle((0.24444649535352503,-3.0840595083809297), 6.679027283956743), linewidth(0.4) + ttqqtt); /* dots and labels */ dot((-5.253612464770877,7.831239053966863),linewidth(4pt) + dotstyle); label("$A$", (-5.091207584201947,8.182886993086111), NE * labelscalefactor); dot((-6.417914253354641,-2.6125144798721553),linewidth(4pt) + dotstyle); label("$B$", (-6.2294076389137505,-2.279798125227352), NE * labelscalefactor); dot((6.902085746645359,-2.432514479872155),linewidth(4pt) + dotstyle); label("$C$", (7.07877761617811,-2.0609134993212126), NE * labelscalefactor); dot((-1.2002582691666517,3.4368479133826457),linewidth(4pt) + dotstyle); label("$O$", (-1.019953542348187,3.8051944749633235), NE * labelscalefactor); dot((-1.1332933334831907,-2.5411006836576764),linewidth(4pt) + dotstyle); label("$K$", (-0.9761766171669638,-2.1922442748648963), NE * labelscalefactor); dot((-16.72710660684508,-2.7518278900544586),linewidth(4pt) + dotstyle); label("$D$", (-16.560761981682433,-2.4111289007710357), NE * labelscalefactor); dot((-3.1925383372509226,2.64276690074704),linewidth(4pt) + dotstyle); label("$M$", (-3.0336921006844553,2.9734328965199937), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $D$ be the intersection of the tangent from $A$ to $\Omega$ and line $\overline{BC}$. We see that $D$ is the inverted image of $M$ with respect to $\omega$, therefore the points $D, M$ and $O$ are collinear. However $\angle OAD = 90^\circ$, therefore $\triangle DMA \sim \triangle DAO \implies DA^2 = DM \cdot DO$. This means that $DA^2 = DB \cdot DC = DM \cdot DO$ which means that the points $B, M, C$ and $O$ are concyclic, as desired.

DapperPeppermint wrote: Kinda easy for Problem 16? Hmm.. very tru. Anyways solution :- Let the tangent to $\omega$ at $A$ meet BC(extended) at $P$. Clearly $PA$ is also tangent to $\Omega$. Also let $M$ be the mid point of $AK$ Spam PoP- $$PM.PO=PA^2=PB.PC$$By Converse of Power Of Point of $P$ with respect to $\Omega$, $BMOC$ is cyclic

Let $M$ be the midpoint of $\overline{AK}$ and let $X=\overline{AA}\cap\overline{KK}\cap\overline{OM},$ which clearly exists. Then, $\overline{XA}$ is tangent to $(AMO),$ meaning $$XB\cdot XC=XA^2=XM\cdot XO.$$Hence, $M$ lies on $(BOC).$ $\square$

Invert about $\omega$, $A$ and $K$ remain fixed and $B,C$ are the intersection of $\odot(OK)$ and a circle tangent to $\omega$ at $A$. Radical centre theorem on $\odot(ABC),\odot(OBKC),\omega$ finishes the problem $\blacksquare$