it is as cute as the second day problem let us suppose DP intersect AB at K KA*KA=KB*KB=KD*KP so K is the middle so KM is the middle line in the triangle ABC let us show that MPC=KPB then the problem will be solved DCB=f(angle)=KMB=KPB=DBK note MPC=x so BPM=k-x (BDC=k) now KPMB cyclic (KMB=KPB) and DPCB cyclic so DBC+DPC=180 but also KBM+KPM=180 so KPB=MPC so we are done

I think too easy for a second round.

Actually if $BD \cap W_1=C'$, then $\triangle APB$, $\triangle BPC$, $\triangle C'PA$ are spirally similar, with spiral similarity center $P$ (apparently).

[asy][asy] import graph; size(16cm); 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 = -14.12, xmax = 7.68, ymin = -8.39, ymax = 7.19; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-8.18,0.53), 3.7643060449437424), linewidth(.7) + wrwrwr); draw(circle((-1.3,0.67), 5.2619388061816155), linewidth(.7) + wrwrwr); draw((xmin, 0.24443676978465095*xmin + 6.404624783463565)--(xmax, 0.24443676978465095*xmax + 6.404624783463565), linewidth(.7) + wrwrwr); /* line */ draw((xmin, -0.22971799065596446*xmin + 2.102230014008526)--(xmax, -0.22971799065596446*xmax + 2.102230014008526), linewidth(.7) + wrwrwr); /* line */ draw((-2.549426614613997,5.781451176984303)--(3.935360294483731,1.1982069546524587), linewidth(.7) + wrwrwr); draw((-5.78,3.43)--(-5.664012332643493,-2.269965367234024), linewidth(.7) + wrwrwr); draw((-5.664012332643493,-2.269965367234024)--(0.692966839934867,3.4898290658183813), linewidth(.7) + wrwrwr); draw((-2.549426614613997,5.781451176984303)--(-5.78,3.43), linewidth(.7) + wrwrwr); draw((-5.664012332643493,-2.269965367234024)--(-2.549426614613997,5.781451176984303), linewidth(.7) + wrwrwr); draw((-5.664012332643493,-2.269965367234024)--(3.935360294483731,1.1982069546524587), linewidth(.7) + wrwrwr); draw((-9.073819622641716,4.186649625296442)--(-5.664012332643493,-2.269965367234024), linewidth(.7) + wrwrwr); draw((-5.811623118627857,4.984050401140373)--(-5.78,3.43), linewidth(.7) + wrwrwr); /* dots and labels */ dot((-8.18,0.53),linewidth(2pt) +dotstyle); label("$W_1$", (-8.1,0.73), NE * labelscalefactor); dot((-5.78,3.43),linewidth(2pt) +dotstyle); label("$D$", (-5.7,3.63), NE * labelscalefactor); dot((-1.3,0.67),linewidth(2pt) +dotstyle); label("$W_2$", (-1.22,0.87), NE * labelscalefactor); dot((-5.664012332643493,-2.269965367234024),linewidth(2pt) + dotstyle); label("$P$", (-5.58,-2.11), NE * labelscalefactor); dot((-2.549426614613997,5.781451176984303),linewidth(2pt) + dotstyle); label("$B$", (-2.46,5.95), NE * labelscalefactor); dot((-9.073819622641716,4.186649625296442),linewidth(2pt) + dotstyle); label("$A$", (-9,4.35), NE * labelscalefactor); dot((3.935360294483731,1.1982069546524587),linewidth(2pt) + dotstyle); label("$C$", (4.02,1.35), NE * labelscalefactor); dot((0.692966839934867,3.4898290658183813),linewidth(2pt) + dotstyle); label("$M$", (0.78,3.65), NE * labelscalefactor); dot((-5.811623118627857,4.984050401140373),linewidth(2pt) + dotstyle); label("$E$", (-5.74,5.15), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $E:=PD\cap AB$. $E$ lies on radical axis of $W_1$ and $W_2$, so $EA=EB$ We have $\angle ABP=\angle BCP$, because tangency. Also, since triangles $EAD$, $EPA$ similar, we have $\angle PBC=\angle PDC=\angle EDA=\angle EAP=\angle BAP$. Thus, $PCB$ and $PBA$, and since $E$ is midpoint of $AB$, same spiral similarity that brings $CB$ to $BA$ also takes $M$ to $E$, implying the conclusion as $\angle BDC=\angle BPC=\angle MPD$.

iran 2010 2nd Round P3 wrote: Circles $W_1,W_2$ meet at $D$and $P$. $A$ and $B$ are on $W_1,W_2$ respectively, such that $AB$ is tangent to $W_1$ and $W_2$. Suppose $D$ is closer than $P$ to the line $AB$. $AD$ meet circle $W_2$ for second time at $C$. Let $M$ be the midpoint of $BC$. Prove that $\angle{DPM}=\angle{BDC}$. Notice that $D$ is the $P-\text{Humpty Point}$ of $\triangle BPA$, so if $PD\cap AB=R$ then $AR=RB$. Hence $RM\|AC$. So, by Homothety at $B$ we get that $\odot(BXM),\odot(BDC)$ are internally tangent at $B$ where $X=BD\cap RM$ and $RB$ is the common tangent to both the circles. Hence, $$RD\cdot RP=RB^2=RX\cdot RM\implies \angle DPM=\angle RXD=\angle BDC. \blacksquare$$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.05015421010951cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -3.9231536695241593, xmax = 12.827103347325025, ymin = -9.04805769080714, ymax = 2.446955861661247; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pair A = (-0.49061490038429356,-0.59975490629623), B = (5.522980152527406,1.1431233059414534), O_1 = (0.44165357896052615,-3.816436987699729), O_2 = (6.7524473745649995,-3.0990078723867294), D = (2.699762430277838,-1.3431572732305677), P = (3.197304676053955,-5.719738109826762), C = (11.163846035069993,-3.3154066762272723), M = (8.343413093798699,-1.0861416851429095); /* draw figures */ draw(A--B, linewidth(0.8)); draw(circle(O_1, 3.3490548114956624), linewidth(0.8)); draw(circle(O_2, 4.416703135168685), linewidth(0.8)); draw(B--C, linewidth(0.8)); draw(A--C, linewidth(0.8)); draw((2.516182626071557,0.27168419982261194)--P, linewidth(0.8)); draw(P--M, linewidth(0.8)); draw(P--C, linewidth(0.8)); draw(D--B, linewidth(0.8)); draw(P--B, linewidth(0.8)); draw(circle((4.945276180299027,-2.486596594688917), 3.6754058873219146), linewidth(2.) + dotted + red); draw(M--(2.516182626071557,0.27168419982261194), linewidth(0.8)); /* dots and labels */ dot(A); label("$A$", (-0.4373972450487933,-0.46671076795747596), NE * labelscalefactor); dot(B); label("$B$", (5.576197807862907,1.2761674442802073), NE * labelscalefactor); dot(O_1); label("$O_1$", (0.320954343482107,-3.5533347774165795), NE * labelscalefactor); dot(O_2); label("$O_2$", (6.800203880579448,-2.9679405687260596), NE * labelscalefactor); dot(D,uuuuuu); label("$D$", (2.7556620750813132,-1.2117579426545009), NE * labelscalefactor,uuuuuu); dot(P,linewidth(4.pt) + uuuuuu); label("$P$", (3.2479253869347047,-5.615518921667274), S* labelscalefactor,uuuuuu); dot(C,uuuuuu); label("$C$", (11.217269273426096,-3.180811190068067), NE * labelscalefactor,uuuuuu); dot(M,uuuuuu); label("$M$", (8.396733540644501,-0.9589740798108675), NE * labelscalefactor,uuuuuu); dot((2.516182626071557,0.27168419982261194),uuuuuu); label("$E$", (2.569400281407057,0.41138054507830346), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] we can see that $BMPE$ is cyclic since $EM||AC$ so: \[\color{green} \angle DPM=\angle BME+\angle BEM =\angle BAC+\angle BCA=\angle BAC+\angle ABD=\angle BDC\] so we are done \[\color{magenta} \huge{\boxed{\mathbf{Q.E.D}}}\]

Let $DP$ meet $AB$ at $S$. $\angle ABP = \angle BCP$ and $BAP = \angle DAP + \angle DPA = \angle PDC = \angle PBC$ so triangles $ABP$ and $BPC$ are similar. we know $S$ is midpoint of $AB$ so $BSP$ and $CMP$ are similar. $\angle DPM = \angle CPB = \angle BDC$. we're Done.