(i) As $\triangle{XA'Y}$ is right, we have $MA'=MX=MY$ and similarly from $\triangle{XB'Y}$ we have $MB'=MX=MY$, giving us our desired result. (ii) Let $D$ and $E$ be the feet of perpendicular from $A$ and $C$ onto $BC$ and $AB$ respectively. The power of $H$ with respect to $k_1$ is $AH\cdot{}HD$ and that of $H$ with respect to $k_2$ is $CH\cdot HE$. Since $AH\cdot HD=CH\cdot HE$, the point $H$ has equal power with respect to $k_1$ and $k_2$. Thus $H$ lies on the radical axis, which is the line $PQ$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -15.387877828827461, xmax = 16.597407099298206, ymin = -8.921681417834973, ymax = 11.063895308497147; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((1.24,8.08)--(-9.86,-1.38)--(6.04,-5.28)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((1.24,8.08)--(-9.86,-1.38), linewidth(2) + zzttqq); draw((-9.86,-1.38)--(6.04,-5.28), linewidth(2) + zzttqq); draw((6.04,-5.28)--(1.24,8.08), linewidth(2) + zzttqq); draw((-2.9304902619207525,-3.0796910678307596)--(3.2078490901584686,2.6028200323922643), linewidth(2)); draw(circle((-0.8452451309603762,2.50015446608462), 5.956754438328758), linewidth(2)); draw(circle((-3.3260754549207654,0.6114100161961322), 6.83065764135454), linewidth(2)); draw(circle((0.1386794141188581,-0.23843551771924765), 4.182407871234052), linewidth(2)); draw((1.24,8.08)--(-2.9304902619207525,-3.0796910678307596), linewidth(2)); draw((-9.86,-1.38)--(3.2078490901584686,2.6028200323922643), linewidth(2)); draw((3.2078490901584686,2.6028200323922643)--(1.5447767266409906,-4.177398065025149), linewidth(2)); draw((-2.9304902619207525,-3.0796910678307596)--(4.314290690614167,-0.47677575554276375), linewidth(2)); draw((xmin, -1.3134811986379942*xmin + 1.7746106206176668)--(xmax, -1.3134811986379942*xmax + 1.7746106206176668), linewidth(2)); /* line */ /* dots and labels */ dot((1.24,8.08),dotstyle); label("$A$", (1.3155487447492764,8.283475115398641), NE * labelscalefactor); dot((-9.86,-1.38),dotstyle); label("$B$", (-9.785226612659043,-1.1657724581391404), NE * labelscalefactor); dot((6.04,-5.28),dotstyle); label("$C$", (6.123794191460978,-5.075085060465612), NE * labelscalefactor); dot((-2.9304902619207525,-3.0796910678307596),dotstyle); label("$X$", (-2.698290932505709,-2.900921901952494), NE * labelscalefactor); dot((3.2078490901584686,2.6028200323922643),dotstyle); label("$Y$", (3.3015631683910662,2.80625638914444), NE * labelscalefactor); dot((0.1386794141188581,-0.23843551771924765),linewidth(4pt) + dotstyle); label("$M$", (0.22846716549271776,-0.07869087888258165), NE * labelscalefactor); dot((4.314290690614167,-0.47677575554276375),linewidth(4pt) + dotstyle); label("$A'$", (4.388644747647625,-0.30865044372531525), NE * labelscalefactor); dot((1.5447767266409906,-4.177398065025149),linewidth(4pt) + dotstyle); label("$B'$", (1.6291299695348223,-4.008908896194756), NE * labelscalefactor); dot((-4.265427545111452,7.377169505274173),linewidth(4pt) + dotstyle); label("$Q$", (-4.182575396490626,7.5517855908990335), NE * labelscalefactor); dot((2.9457360371257666,-2.0945582802974196),linewidth(4pt) + dotstyle); label("$P$", (3.0297927735769266,-1.9183673976244504), NE * labelscalefactor); dot((-0.23189443687275468,2.0791996035187688),linewidth(4pt) + dotstyle); label("$H$", (-0.14783030424993715,2.241810184530457), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Part 1: We have that $90 = \angle AA'X = \angle YA'X$, also we have that $90 = \angle BB'Y = \angle XB'Y$, thus we have that $XB'A'Y$ is a cyclic quad, from here we have that $MA'=MB'$ Part 2: We see that the foot from the $A$-altitude belongs to the $k_1$ (we call the point $D$), and that the foot from the $B$-altitutde belongs to $k_2$ (and we call this one $E$). We now draw in the circle with center $T$, where $T$ is the midpoint of $AB$, now with respect to that circle we have that the power of point $H$ is $AH.HD=BH.HE$, thus we have that $H$ belongs to the radical axis of $k_1$ and $k_2$, which is $\overline{PQ}$