Let $K'$ on the circle with diameter $AB$ such that $CK'$ is tangent to the circle with diameter $AB$ and $K',A$ are on the same side of line $CC_1$. Let $L'=BK'\cap CC_1$ Because $AK'L'C_1$ is cyclic then we have $\angle BK'C=\angle BAK'=\angle K'L'C\Rightarrow \angle L'K'H+\angle HK'C=\angle L'K'C_1+\angle L'C_1K'$ Buts $CK'^2=CB_1.CA=CH.CC_1\Rightarrow \Delta CK'H\sim \Delta CC_1K'\Rightarrow \angle HK'C=\angle L'C_1K'$ Hence $\angle L'K'H=\angle L'K'C_1\Rightarrow K\equiv K',L\equiv L'(?)\Rightarrow CK=CL=CM$ We have $CM^2=CK^2=CA_1.CB\Rightarrow \angle CMB=90^0\Rightarrow BM^2=BA_1.BC=BC_1.BA$ But if we let $P'$ and $Q'$ be intersections of the line $CC_1$ and the circle with diameter $AB$ then it's easy to prove that $BP'^2=BQ'^2=BC_1.BA$. Thus $BM^2=BP'^2=BQ'^2$ so $P\equiv P',Q\equiv Q'$ and the problem is done.

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Let $K'$ be the tangency point from C to $(AB_1A_1B)$, and redefine L' accordingly. Then, we have $\angle B_1KB=\angle A=180-\angle B_1HL'$, which means $B_1K'L'H$ is cyclic. Also, we have $\angle AK'B=90=AC_1L'$, so $AK'L'C_1$ is cyclic. Then, we have $\angle CK'B=\angle K'AB=\angle K'L'C$, which means $K'C=L'C$. Let the circles centered at $C$ and $B$ defined in the problem be $\omega_C$ and $\omega_B$, respectively. Now, notice that we have $CL'^2=CK'^2=CB_1\cdot CA$, so $\angle CB_1L'=\angle CL'A$. We know $\angle CB_1L'=90+\angle BB_1L'=90+\angle HK'L'$ and $\angle CL'A=180-\angle AL'C_1=90+\angle L'AC_1=90+\angle L'K'C_1$, so we know that $\angle HK'L'=\angle C_1K'L'$, so $L'=L$ and $K'=K$. Then, let $T:=A_1B_1\cup AB$, and $X$ the midpoint of $AB$, and we know that $HT\perp CX$ by Brokard on $ABA_1B_1$. It's well known that the circle with diameter AB is an apollonius circle of C and T, so $\angle TKA=\angle C_1KA$. Thus, we have $\angle TKH=2\cdot AKC_1+2\cdot BKC=180$, so K lies on HT. Thus, we will have proved that $H$ is the radical center of $\omega_C$, $\omega_B$, and the circle with diameter AB, because the radical axis of $\omega_C$ and $\omega_B$ is $AA'$, which contains $M$ which is defined to be on both circles; the radical axis of $\omega_C$ and $(AB)$ is $KH$, because the line connecting the center of their circles is $XC$, which is perpendicular to $KH$, and $K$ lies on both circles. This means that $H$ lies on the radical axis of $\omega_B$ and $(AB)$, which means that the intersections between $CC_1$ and $\omega_B$, or $P$ and $Q$, both lie on $(AB)$, which shows that $A,K,P,Q$ lie on the same circle, as desired.

dang ur pretty smart

See that it suffices to prove that $|BM|^2=|BC_1|\cdot |BA|$ (This gives us $|BQ|^2=|BP|^2=|BM|^2=|BC_1|\cdot |BA|\Rightarrow \angle AQB=\angle APB=90^\circ\Rightarrow AKPQ$ is cyclic.) Now, we can get rid of the points $P$ and $Q$. Also, $|BM|^2=|BC_1|\cdot |BA|\Leftrightarrow \angle C_1MB=\angle BAM(=\angle 90^\circ-\angle ABC=\angle C_1CB)\Leftrightarrow MC_1A_1C\text{ is cyclic. }\Leftrightarrow \angle BMC=90^\circ$. Thus, it suffices to prove that $\angle BMC=90^\circ$. Let $KH\cap AB=T$. We have $\angle AKB=90^\circ$ and $\angle C_1KB=\angle BKT$. Hence, $(A,B;C_1,T)=-1$. Also, $(A,B;C_1,AB\cap B_1A_1)=-1$. Hence, $AB\cap B_1A_1=T\Rightarrow B_1A_1$ passes through $T$. Using Brocard's Theorem, one can get that polar of $C$ wrt $(ABA_1B_1)$ is $TH$. Hence, $CK$ is tangent to $(ABA_1B_1)$. Let $AK\cap C_1C=S$. We have $\angle BC_1S=90^\circ=\angle BKS\Rightarrow BC_1KS$ is cyclic. Then, $\angle KSC=\angle KBC_1=90^\circ-\angle KAB=90^\circ-\angle LKC=\angle SKC\Rightarrow |KC|=|CS|$ and $\angle LKS=90^\circ$. Hence, $|KC|=|CS|=|CL|$. Then, $|CM|^2=|CL|^2=|CK|^2=|CA_1|\cdot |CB|$ and $\angle MA_1C=90^\circ$. Thus, $\angle BMC=90^\circ$, as desired.

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