As shown in the figure below, $ABCD$ is a trapezoid, $AB \parallel CD$. The sides $DA$, $AB$, $BC$ are tangent to $\odot O_1$ and $AB$ touches $\odot O_1$ at $P$. The sides $BC$, $CD$, $DA$ are tangent to $\odot O_2$, and $CD$ touches $\odot O_2$ at $Q$. Prove that the lines $AC$, $BD$, $PQ$ meet at the same point. [asy][asy] size(200); defaultpen(linewidth(0.8)+fontsize(10pt)); pair A=origin,B=(1,-7),C=(30,-15),D=(26,6); pair bisA=bisectorpoint(B,A,D),bisB=bisectorpoint(A,B,C),bisC=bisectorpoint(B,C,D),bisD=bisectorpoint(C,D,A); path bA=A--(bisA+100*(bisA-A)),bB=B--(bisB+100*(bisB-B)),bC=C--(bisC+100*(bisC-C)),bD=D--(bisD+100*(bisD-D)); pair O1=intersectionpoint(bA,bB),O2=intersectionpoint(bC,bD); dot(O1^^O2,linewidth(2)); pair h1=foot(O1,A,B),h2=foot(O2,C,D); real r1=abs(O1-h1),r2=abs(O2-h2); draw(circle(O1,r1)^^circle(O2,r2)); draw(A--B--C--D--cycle); draw(A--C^^B--D^^h1--h2); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,dir(350)); label("$D$",D,dir(350)); label("$P$",h1,dir(200)); label("$Q$",h2,dir(350)); label("$O_1$",O1,dir(150)); label("$O_2$",O2,dir(300)); [/asy][/asy]
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