how would you possibly do this in a semi-elegant synthetic way

khina wrote: how would you possibly do this in a semi-elegant synthetic way Solution 1: Use cartesian coordinates with the origin in $C$ and axes $CA$, $CP$. The complete solution will be usually $1$ page. Solution 2: (following the official solution) We have to use the converse of Menelaus' theorem in $\triangle BCD$. Do a few natural calculations and the result yields.

WOW I overcomplicated this a lot - does this proof even work (this is essentially the converse)? Let M be the intersection of AC and BD, and E be the midpoint of BC. Also say X is on BC such that AB^2 = BC*BX, and let Y be the intersection of AX and CF. We shall prove that the circumcenter P of CXY satisfies that AP and BD are perpendicular and AC and CP are perpendicular. First of all, note that the condition gives us that XBA is similar to ABC and XCY. This gives us that $\angle{PCA} = 180 - \angle{PCX} - \angle{ACB} = 90 + \angle{CYX} - \angle{ACB} = 90$ which gives us that AC and CP are perpendicular. Now, let F be the foot of the perpendicular dropped from P to AD, and let N be the foot of the perpendicular dropped from P to BD. Note that DFPN, MCPN, and AFPC are all cyclic. Angle chasing now gives us that $\angle{AFC} = \angle{APC} = \angle{AMN} = \angle{BMC}$, and $\angle{FAC} = \angle{MCB}$, so FAC and MCB are similar. Thus, $\angle{AFM} = \angle{CME} = \angle{BAC} = \angle{MCD}$, so MCFD is cyclic. Now, by radical axis theorem, DF, NP, and MC must concur at a point (which must be A), so A lies on NP and AP is perpendicular to BD, as desired.

Sorry but I can't construct this. Is it possible for someone to post an image.

We denote by $\Gamma$ the circumcircle of our circle. First of all because $PC \perp AC$ we have that $AC$ is tangent line to $\Gamma$. Let $AZ$ be the other tangent line to $\Gamma$. So we must have that $PA \perp ZC$, so $DB \parallel ZC$. Now we use harmonic division. $(X,C;Y,Z) \stackrel{C}{=} (B,DB \cap AC, D, P_{\infty}) $, but $DB \cap AC$ is the midpoint of $DB$, so it's a harmonic divison, so $A-Y-X$ are colliniar. [asy][asy] import graph; size(8cm); 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 = -12.2, xmax = 12.2, ymin = -7.25, ymax = 7.25; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-4.039743127062307,0.39235535461758214)--(-3.711136975264292,0.1239936639825363)--(-3.4427752846292465,0.45259981578055125)--(-3.7713814364272613,0.7209615064155972)--cycle, linewidth(0.6)); draw((-0.4471684078444506,2.5348300714317697)--(-0.7919984792762205,2.2876616635873193)--(-0.54483007143177,1.9428315921555495)--(-0.2,2.19)--cycle, linewidth(0.6)); draw((-1.8911245634895693,3.0233168610331784)--(-1.5625184116915543,2.7549551703981328)--(-1.2941567210565084,3.083561322196148)--(-1.6227628728545236,3.3519230128311936)--cycle, linewidth(0.6)); /* draw figures */ draw((-5.92,-1.91)--(-1.26,-1.33), linewidth(0.6) + wrwrwr); draw((-5.92,-1.91)--(-0.2,2.19), linewidth(0.6) + wrwrwr); draw((-4.86,1.61)--(-1.26,-1.33), linewidth(0.6) + wrwrwr); draw(circle((-1.3085923842821308,3.7366215702667773), 1.9028965699960712), linewidth(0.6) + wrwrwr); draw((-5.92,-1.91)--(0.46969661092200743,4.413898179665534), linewidth(1.6) + dotted + wrwrwr); draw((-3.045525745709047,4.513846025662389)--(-0.2,2.19), linewidth(0.6) + wrwrwr); draw((-3.045525745709047,4.513846025662389)--(-5.92,-1.91), linewidth(0.6) + wrwrwr); draw((-4.86,1.61)--(-5.92,-1.91), linewidth(0.6) + wrwrwr); draw((-5.92,-1.91)--(-1.3085923842821308,3.7366215702667773), linewidth(0.6) + wrwrwr); draw((0.46969661092200743,4.413898179665534)--(-1.26,-1.33), linewidth(0.6) + wrwrwr); draw((-4.86,1.61)--(-0.2,2.19), linewidth(0.6) + wrwrwr); draw((-1.3085923842821308,3.7366215702667773)--(-0.2,2.19), linewidth(0.6) + wrwrwr); draw((-1.3085923842821308,3.7366215702667773)--(-3.045525745709047,4.513846025662389), linewidth(0.6) + wrwrwr); /* dots and labels */ dot((-5.92,-1.91),dotstyle); label("$A$", (-6.32,-1.87), NE * labelscalefactor); dot((-1.26,-1.33),dotstyle); label("$B$", (-1.08,-1.63), NE * labelscalefactor); dot((-0.2,2.19),dotstyle); label("$C$", (0.06,1.93), NE * labelscalefactor); dot((-4.86,1.61),linewidth(4pt) + dotstyle); label("$D$", (-5.22,1.81), NE * labelscalefactor); dot((-1.3085923842821308,3.7366215702667773),linewidth(4pt) + dotstyle); label("$P$", (-1.34,4.11), NE * labelscalefactor); dot((-2.004239071286051,1.9654380555051696),linewidth(4pt) + dotstyle); label("$Y$", (-2.06,1.59), NE * labelscalefactor); dot((0.46969661092200743,4.413898179665534),linewidth(4pt) + dotstyle); label("$X$", (0.68,4.29), NE * labelscalefactor); dot((-3.045525745709047,4.513846025662389),linewidth(4pt) + dotstyle); label("$Z$", (-3.6,4.47), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

khina wrote: ... Angle chasing now gives us that $\angle{AFC} = \angle{APC} = \angle{AMN} = \angle{BMC}$, and $\angle{FAC} = \angle{MCB}$, so FAC and MCB are similar. ... We don't know that $\angle{APC}$ is equal to $\angle{AMN}$! $\angle{NPC}$ is equal to $\angle{AMN}$ because of the cyclic quadrilateral $MCPN$. But if you assume that $\angle{APC}=\angle {NPC}$, then you assume that $A-N-P$ are collinear which is what you need to prove.

The following claim kills the problem. Claim. $\triangle ADY \sim \triangle ADC$. Proof. We will bash this, assume that $ABCD$ is not a rectangle since that's easier to bash. WLOG let the coordinates be $A(c,b),B(1,b),C(0,0),D(a,0)$ with $a,b\neq 0$. Because $BC \parallel AD$, then $$\frac{b-0}{1-0}=\frac{b-0}{c-a} \Rightarrow c=a+1$$The coordinates now are $A(a+1,b),B(1,b),C(0,0),D(a,0)$. The equation of line $AP$ is $$\frac{a-1}{b}(x-(a+1))=y-b\Rightarrow (a-1)x-by-a^2+b^2+1=0$$. The equation of line $CP$ is $$-\frac{a+1}{b}(x-0)=y-0 \Rightarrow (a+1)x+by=0$$Clearly $P$ is the intersection of line $AP$ and $CP$, therefore by some calculations we got $P\left(\frac{a^2-b^2-1}{2a},-\frac{(a+1)(a^2-b^2-1)}{2ab}\right)$. Then because $PY=PC$, we have $Y\left(\frac{a^2-b^2-1}{a},0 \right)$. Therefore, $$DY\cdot DC=\left|\frac{(b^2+1)}{a}\right|\cdot |a|=b^2+1=DY^2$$Since $\angle ADY = \angle ADC$, the claim is proved. Now, let the tangent from $X$ to $(PYC)$ intersect $AC$ at $M$ and $CP$ intersect $(PYC)$ at $N$. Then, since $AC$ is tangent to $(PYC)$ $$\angle XYC = \angle XNC = \angle XCM= \angle ACB=\angle CAD=\angle DYA$$Because $D,Y,C$ are collinear, furthermore $A$ and $X$ are not on the same side divided by $CD$. Then, $A,Y,X$ are collinear.