This is not a beautiful geo at all ,...

This looks beautiful. Here's a solution using Miquel points. Let the diagonals meet at $E$. Let the point where $(DEA)$ meet $(CBE)$ be $P$. Then since $\angle PDX = \angle PDA - \angle XDA = \angle PBC - \angle XBC = \angle PBX$, $B, D, X, P$ are concyclic. Define $Q$ similarly. Then we have $\angle AXC = \angle AQC = \angle AQE + \angle EQC = \angle ABD + 180 - \angle BDC$, and $\angle BXD = \angle BCA + 180 - \angle DAC$, so $X, B, D, AD \cap BC$ and $X, A, C, AB \cap CD$ are sets of concyclic points. Note that due to inversion under the $B-$Apollonian circle of $ABC$, the line $BD$ subtends equal angles at $AB \cap CD$, and the inverse of $AD \cap BC$. By angle chasing, we're done.

An elementary solution does exist?

After inverting around X it doesn’t look bad at all

The below solution uses two spiral similarities and an inversion. No diagram but I'll assume $AB\cap CD, BC\cap DA$ both lie on the same side of line $BD$ as point $C$. Let $E=AB\cap CD, F = AD\cap BC$ so trivially $X = (ACE) \cap (BDF)$. Construct $X'$ with $X'CD \sim XBA$ so $\angle X'DC = \angle XAB = \angle XCD \implies XC||X'D$. Now we'd like to show $\angle AXB+\angle CXD = \angle DXC + \angle DX'C = 180^{\circ}$, or equivalently that $XCX'D$ is an isosceles trapezoid. If $XCX'D$ is a parallelogram it's not hard to show $ABCD$ is a kite symmetric about $AC$ and we can easily show $\angle AXB=\angle CXD = 90^{\circ}$; otherwise, if $XCX'D$ is not a parallelogram, it's enough to show $X'C = XD$. Since $X'C = BX \frac{CD}{AB}$ by similarity, we want to show $\dfrac{BX}{DX} = \dfrac{AB}{CD}$. Construct $P=(ACE)\cap (BDE)$ so $PAB\sim PCD$. Then $\dfrac{AB}{CD} = \dfrac{PB}{PD}$. Let $\omega$ be the circle centered at point $O$ along $AC$ passing through $B,D$; this is the Apollonius circle of $AC$ through $B,D$ which exists because $AB\cdot CD = AD \cdot BC$. I claim that $P, X$ are inverses in $\omega$, which will imply $\dfrac{PB}{PD}=\dfrac{XB}{XD}$, thus finishing the problem. Since $P,X$ both lie on $(ACE)$, which is fixed in inversion about $\omega$, and $P,X$ also lie on $(BDE), (BDF)$ respectively, it's enough to show $(BDE), (BDF)$ are inverses in $\omega$. This is directly equivalent with $\angle BED+\angle BFD = \angle BOD$. But $\angle BED +\angle BFD = (180^{\circ} - \angle A - \angle D) + (180^{\circ} - \angle A - \angle B) = \angle C - \angle A$, Meanwhile, $\angle BOD = \angle BOC+\angle DOC = (\angle BCA - \angle CBO) + (\angle DCA - \angle CDO) = \angle BCA +-\angle CAB + \angle DCA - \angle CAD = \angle C - \angle A$, so the equalities hold and we're done.

I think that I don't use $AB.CD=BC.DA$. A convex quadrilateral $ABCD$ and point $X$ lies inside $ABCD$ so that $\angle{XAB} = \angle{XCD}$ and $\angle{XBC} = \angle{XDA}$.

This looks like some old problems, but I cannot recall where I saw the similar problem. Maybe it's new I guess. Lemma. Suppose that $ABCD$ and $A'B'C'D'$ are quadrilaterals such that they share the same angles and $AB\cdot CD=BC\cdot DA$, $A'B'\cdot C'D'=B'C'\cdot D'A'$, then $ABCD\sim A'B'C'D'$.

Now take inversion w.r.t. $X$, and let $A_0,B_0,C_0,D_0$ be the images of $A,B,C,D$ under this inversion. Then we have \[A_0B_0\cdot C_0D_0 = \frac{r^4}{XA\cdot XB\cdot XC\cdot XD}(AB\cdot CD) = \frac{r^4}{XA\cdot XB\cdot XC\cdot XD}(BC\cdot DA) = B_0C_0\cdot D_0A_0. \]Moreover, we have $\angle A_0B_0C_0 = \angle A_0B_0X+\angle XB_0C_0 = \angle XAB+\angle BCX =\angle BCX+\angle XCD = \angle BCD$. Similarly we have $\angle C_0 = \angle D, \angle D_0 = \angle A, \angle A_0=\angle B$. Therefore by the lemma we have $ABCD\sim D_0A_0B_0C_0$. Now by scaling we can assume that $D_0A_0B_0C_0\cong ABCD$, and so there is an isometry $f$ such that $f(D_0)=A$ and etc. It is clear that $f(X)$ is the isogonal conjugate of $X$, and so it is well-known that $\angle AXB+\angle CXD = 180^{\circ}$.

USJL wrote: This looks like some old problems, but I cannot recall where I saw the similar problem. Maybe it's new I guess. I saw something like this before too..

This one is much harder than Problem 1..., I haven't done yet...

Construct $E$ such that $\triangle XCD$ spirals similarly to $\triangle XBE$. Then we also have $\triangle XBC\sim\triangle XED$. The condition $AB\cdot CD=AC\cdot BD$ is equivalent to $\frac{AB\cdot BE}{BX}=\frac{AD\cdot DE}{DX}$, and it suffices to show $A,X,E$ are collinear. Suppose otherwise, so WLOG assume $\angle AXB+\angle BXE>180^\circ$. We claim that $\frac{AB\cdot BE}{BX}>AE>\frac{AD\cdot DE}{DX}$. Extend $AX$ to intersect circumcircle of $\triangle BXE$ at $P$, then $\triangle ABP\sim\triangle BXE$. Hence $\frac{AB\cdot BE}{BX}=AP$. Similarly extend $EX$ to intersect circumcircle of $\triangle AXD$ at $Q$, then $\frac{AD\cdot DE}{DX}=EQ$. Now we fix $A,X,E$ and vary $B,D$, ensuring $\angle XAB=\angle XBE$ and $\angle XED=\angle XDA$. Since $\angle APE=\angle EBX$, $AP$ is minimal iff $\angle XAB$ is maximal. But $\angle XAB$ is maximal iff $AB$ is tangent to the circumcircle of $\triangle BXE$, which happens when $\triangle AXB\sim\triangle BXE$. Similarly $EQ$ is maximal when $\angle XDA$ is maximal when $\triangle AXD\sim\triangle DXE$. Now it suffices to consider these extreme cases. In this case, $ABED$ becomes cyclic and harmonic, and $X$ is the midpoint of $BD$, so $BE>ED$. Now angle chasing gives $\angle AEP=180^\circ-\angle BDE$, so $\frac{AP}{AE}=\frac{\sin\angle AEP}{\sin\angle APE}=\frac{\sin\angle BDE}{\sin\angle DBE}=\frac{BE}{ED}>1$. Similarly $\frac{EQ}{EA}=\frac{AD}{AB}=\frac{DE}{EB}<1$, hence $AP>AE>EQ$ and we are done. OR alternatively we can fix $A,B,E$ and vary $X$ ensuring $\angle XAB=\angle XBE$, and notice that $BX$ is maximal when $X$ is on $AE$, in which case $\frac{AB\cdot BE}{BX}=AE$, and similarly to show $AE>\frac{AD\cdot DE}{DX}$.

this is not very rigorous answer. point K make $\triangle XKC \sim \triangle ADX$, so $\angle XKC = \angle ADX = \angle XBC$, $B,K,C,X$ are on a circle. from Ptolemy theorem, $BK\times XC+BX\times KC=XK\times BC$ on another hand, form similar we have $\frac{XK}{AD}=\frac{KC}{DX}=\frac{XC}{AX}$ so, $BX\times DX+AX\times KB=AD\times BC$ in like manner, point M make $\triangle AXB \sim \triangle MDX$, so $CX\times AX+BX\times MC=AB\times DC$, so $BX\times DX+AX\times KB=CX\times AX+BX\times MC$, that is $BX\times (CM-DX)+AX\times (XC-BK)=0$ if $\angle AXB+\angle DXC > 180^{\circ}$, then $\angle MDX+\angle DXC > 180^{\circ}$, that is $CM>DX$ and $\angle AXD+\angle BXC < 180^{\circ}$, then $\angle XCK+\angle BXC < 180^{\circ}$, that is $XC>BK$ then, $BX\times (CM-DX)+AX\times (XC-BK)>0$, we get a contradiction! so, $\angle AXB+\angle DXC = 180^{\circ}$

Here's a solution with simple angle chasing. Let $\odot(O)$ be the circle passes through $B,D$ and $O,A,C$ are collinear. By $AB\cdot CD=BC\cdot DA$, $\odot(O)$ is the Apollonian circle of $B,D$ that passes through $A,C$. By $\angle{XAB} = \angle{XCD}$ and $\angle{XBC} = \angle{XDA}$, $X\in\odot(ACE)\cap \odot(BDF)$. Let $E'$ be the image of $E$ under the inversion w.r.t. $\odot(O)$, since $\measuredangle BE'D=\measuredangle EBO+\measuredangle ODE=\measuredangle BCO+\measuredangle OAD=\measuredangle BFD$, $E’\in\odot(BDF)$. Then $\measuredangle BXA=\measuredangle BXE'+\measuredangle E'XA=\measuredangle BDE'+\measuredangle OEA=\measuredangle BDO+\measuredangle CEO+\measuredangle OEA=\measuredangle BDO+\measuredangle CEA$. Similarly, $\measuredangle DXC=\measuredangle DBO+\measuredangle AEC$, and hence $\angle{BXA} + \angle{DXC} = 180$.

Li4 wrote: Here's a solution with simple angle chasing. Let $\odot(O)$ be the circle passes through $B,D$ and $O,A,C$ are collinear. By $AB\cdot CD=BC\cdot DA$, the image of $A$ under inversion w.r.t. $\odot(O)$ is $C$. By $\angle{XAB} = \angle{XCD}$ and $\angle{XBC} = \angle{XDA}$, $X\in\odot(ACE)\cap \odot(BDF)$. Let $E'$ be the image of $E$ under the inversion w.r.t. $\odot(O)$, since $\measuredangle BE'D=\measuredangle EBO+\measuredangle ODE=\measuredangle BCO+\measuredangle OAD=\measuredangle BFD$, so the image of $\odot(BDE)$ under the inversion w.r.t. $\odot(O)$ is $\odot(BDF)$. Then $\measuredangle BXA=\measuredangle BXE'+\measuredangle E'XA=\measuredangle BDE'+\measuredangle OEA=\measuredangle BDO+\measuredangle CEO+\measuredangle OEA=\measuredangle BDO+\measuredangle CEA$. Similarly, $\measuredangle DXC=\measuredangle DBO+\measuredangle AEC$, and hence $\angle{BXA} + \angle{DXC} = 180$. But A，B，C，D may not share a circle. Then the image of A under inversion WRT $\odot(O)$ may not be C.

This problem was proposed by me.

timon92 wrote: This problem was proposed by me. Any more about the background ? Nice problem.

Gao-JX wrote: Li4 wrote: Here's a solution with simple angle chasing. Let $\odot(O)$ be the circle passes through $B,D$ and $O,A,C$ are collinear. By $AB\cdot CD=BC\cdot DA$, the image of $A$ under inversion w.r.t. $\odot(O)$ is $C$. By $\angle{XAB} = \angle{XCD}$ and $\angle{XBC} = \angle{XDA}$, $X\in\odot(ACE)\cap \odot(BDF)$. Let $E'$ be the image of $E$ under the inversion w.r.t. $\odot(O)$, since $\measuredangle BE'D=\measuredangle EBO+\measuredangle ODE=\measuredangle BCO+\measuredangle OAD=\measuredangle BFD$, so the image of $\odot(BDE)$ under the inversion w.r.t. $\odot(O)$ is $\odot(BDF)$. Then $\measuredangle BXA=\measuredangle BXE'+\measuredangle E'XA=\measuredangle BDE'+\measuredangle OEA=\measuredangle BDO+\measuredangle CEO+\measuredangle OEA=\measuredangle BDO+\measuredangle CEA$. Similarly, $\measuredangle DXC=\measuredangle DBO+\measuredangle AEC$, and hence $\angle{BXA} + \angle{DXC} = 180$. But A，B，C，D may not share a circle. Then the image of A under inversion WRT $\odot(O)$ may not be C. The circle $(O)$ is the Apollonian circle of $A,C$ that passes through $B$. So it also passes through $D$

I don't using the transformation We can solve the problem by using the nice Lemma: A convex quadrilateral $ABCD$ satisfies $AB\cdot CD$ = $BC\cdot DA$. Point $K$ lies on the segment $BD$ such that: $\widehat{KAD}=\widehat{CAB}$. Then prove: $\widehat{AKB}=\widehat{CKB}$ The Lemma can be proved easily by using the Law of Sines Now we note that the point $X$ is unique. So Let $X'$ be the second intersection of $(AKB)$ and $(CKD)$ we have: $\widehat{X'AB}=\widehat{X'KB}=\widehat{X'CD}$ Let $(AKB) \cap BC = \{B,U\}$ and $(CKD) \cap AD = \{D,V\}$ We have: $\widehat{AUB}=\widehat{AKB}=\widehat{CKB}=\widehat{CVA}$ so $AUCV$ is cyclic So: $\widehat{KUC}=\widehat{BAK}=\widehat{DAC}={VUC}$ so $\overline{U,K,V}$ From that:$\widehat{X'BC}=\widehat{UKX'}=\widehat{X'DA}$. So $X \equiv X'$ and $\widehat{BXA} + \widehat{DXC} = 180$.

@above i believe that X is not unique since for example there are 2 Brokkard points of harmonic quadrilateral

PcelicaMaja wrote: @above i believe that X is not unique since for example there are 2 Brokkard points of harmonic quadrilateral Oh, $X$ lies insides $ABCD$ and We can calculate $\widehat{AXC}, \widehat{BXD}$ so $X$ is unique I got it when I construct $X$

Invert at $X$ with unit radius, images denote with hatches. Claim. $ABCD\stackrel{+}{\sim} D'A'B'C'.$ Proof. Let spiral similarity $D'A'\mapsto AB$ maps $B'C'\mapsto C''D''.$ Observe that $$|A'B'|\cdot |C'D'|=\frac{|AB|\cdot |CD|}{|XA|\cdot |XB|\cdot |XC|\cdot |XD|}=|A'D'|\cdot |B'C'|,$$also $\measuredangle BAD=\measuredangle BAX+\measuredangle XAD=\measuredangle XD'C'+\measuredangle A'D'X=\measuredangle A'D'C'$ and all similar equalities hold. Thus $B''\in BC,C''\in AD,B''C''\parallel CD,$ and moreover $$\frac{|AB|}{|BB''|}=\frac{|AC''|}{|B''C''|}\implies \frac{|BC|}{|BB''|}\cdot \frac{|AD|}{|AC''|}=\frac{|CD|}{|B''C''|}\stackrel{\text{Thales}}{\implies} B''=C,C''=D\text{ } \Box$$ Now let $ABCDX'\stackrel{+}{\sim} D'A'B'C'X.$ Since $\measuredangle BAX'=\measuredangle A'D'X=\measuredangle XAD$ and all similar arguments hold, $X'$ is the isogonal conjugate of $X$ wrt $ABCD.$ The conclusion follows.

USJL wrote: This looks like some old problems, but I cannot recall where I saw the similar problem. Maybe it's new I guess. Lemma. Suppose that $ABCD$ and $A'B'C'D'$ are quadrilaterals such that they share the same angles and $AB\cdot CD=BC\cdot DA$, $A'B'\cdot C'D'=B'C'\cdot D'A'$, then $ABCD\sim A'B'C'D'$.

Now take inversion w.r.t. $X$, and let $A_0,B_0,C_0,D_0$ be the images of $A,B,C,D$ under this inversion. Then we have \[A_0B_0\cdot C_0D_0 = \frac{r^4}{XA\cdot XB\cdot XC\cdot XD}(AB\cdot CD) = \frac{r^4}{XA\cdot XB\cdot XC\cdot XD}(BC\cdot DA) = B_0C_0\cdot D_0A_0. \]Moreover, we have $\angle A_0B_0C_0 = \angle A_0B_0X+\angle XB_0C_0 = \angle XAB+\angle BCX =\angle BCX+\angle XCD = \angle BCD$. Similarly we have $\angle C_0 = \angle D, \angle D_0 = \angle A, \angle A_0=\angle B$. Therefore by the lemma we have $ABCD\sim D_0A_0B_0C_0$. Now by scaling we can assume that $D_0A_0B_0C_0\cong ABCD$, and so there is an isometry $f$ such that $f(D_0)=A$ and etc. It is clear that $f(X)$ is the isogonal conjugate of $X$, and so it is well-known that $\angle AXB+\angle CXD = 180^{\circ}$. There is a similar problem in ARMO (all-russian, 3 round)... I guess it was in 2011 (or maybe, 2010). It asks to show that $\angle BAC+\angle CBD +\angle DCA +\angle ADB = 180^{\circ}$ under the condition $AB\cdot CD = AD\cdot BC$. Its proof also uses the Apollonian circles, isogonal conjugates and many the same approaches stated above (to solve this IMO question). look at 11.8 in https://olympiads.mccme.ru/vmo/2012/iii-day2.pdf

Clawson Schmidt Conjugates ftw! Claim 1 : Point $X$ is unique. Proof : Let $P = AD \cap BC$ and $Q = AB\cap CD$. Note $\angle XAB = \angle XCD \implies XACQ$ cyclic. Similarly $XBDP$ cyclic. Hence, $X = (ACQ) \cap (BDP)$, and clearly unique such point lies inside convex quadrilateral $ABCD$. $\square$ We will now characterize $X$. Let $AC\cap BD = E$. Define the following two points : $Y$ on $BD$ such that $\frac{BE}{ED} \cdot \frac{BY}{YD} = \left(\frac{BA}{AD}\right)^2=\left(\frac{BC}{CD}\right)^2$ $Z$ on $AC$ such that $\frac{AE}{EC} \cdot \frac{AZ}{ZC} = \left(\frac{AB}{BC}\right)^2=\left(\frac{AD}{DC}\right)^2$ Claim 2 : $(ABY),(CDY),(ADZ),(BCZ)$ concur, and $Y$ and $Z$ are isogonal conjugates in $ABCD$. Proof : By the ratio condition, we must have $AE$ and $AY$ are isogonal in $\angle BAD$. Hence $AX$ and $AY$ are isogonal in $\angle BAD$. We similarly get for $B,C,D$; and we infer that $Y$ and $Z$ are isogonal conjugates in quadrilateral $ABCD$. As they are isogonal conjugates, they are clawson schmidt conjugates in quadrilateral $ABCD$ too! Hence, $(ABY), (CDY), (ADZ),(BCZ)$ concur. $\square$ Call the concurrency point $X$. Claim 3 : $\angle XAB = \angle XCD$ and $\angle XBC = \angle XDA$. Proof : Using the last claim, $\angle XAB = \angle XYB = 180 - \angle XYD =\angle XCD$. We similarly get $\angle XBC = \angle XDA$. $\square$ Claim 4 : $\angle AXB + \angle CXD = 180$. Proof : Note $\angle AXB = \angle AYB$ and $\angle CXD = \angle CYD$. And $Y$ has a isogonal conjugate in $ABCD$, we must have that $\angle AYB + \angle CYD = 180$. $\square$. Combining claims 1,3,4 we are done!

LeVietAn wrote: I think that I don't use $AB.CD=BC.DA$. A convex quadrilateral $ABCD$ and point $X$ lies inside $ABCD$ so that $\angle{XAB} = \angle{XCD}$ and $\angle{XBC} = \angle{XDA}$. Can someone point out the mistake here

... SO. Apollonian circles are NOT ded. anyways angles are taken directed but not written in the correct order of (counter)clockwise stuff (essentially mod 180 for config issues but not written in the write order because i don't wanna) By the length condition, there's a point O on AC which is the center of Apollonian circles B of ABC and D of ADC (which coincide); call this circle w, and call the well-known fact of w that OB,OD are tangents to (ABC),(ADC) fact (1)). Let $E=AB\cap CD,F=AD\cap BC:\measuredangle XAB=\measuredangle XCD=-\measuredangle XCE$, so AXCE is cyclic and analogously so is BXDF. Letting G be the inverse of E wrt w, then $OE\cdot OG=OB^2\stackrel{(1)}{=}OA\cdot OC\implies G\in(AXCE)$ and $$\measuredangle BGD=\measuredangle OGB-\measuredangle OGD=\measuredangle OBE-\measuredangle ODE=-\measuredangle OBA-\measuredangle ODC\stackrel{(1)}{=}-\measuredangle OCB-\measuredangle OAD=\measuredangle ACB-\measuredangle CAF=\measuredangle CFA=\measuredangle BFD\implies G\in(BXDF).$$The problem is bashy angle chasing from here: $$\measuredangle AXB=\measuredangle AXG-\measuredangle BXG=-\measuredangle OEB-\measuredangle BDG=-\measuredangle OEB-(\measuredangle BDO-\measuredangle ODG)=-\measuredangle ODB+\measuredangle ODG-\measuredangle OBG$$$$\measuredangle CXD=\measuredangle GXD-\measuredangle GXC=\measuredangle GBD-\measuredangle OED=\measuredangle OBD+\measuredangle OBG-\measuredangle ODG,$$so adding them together immediately implies the result. Remark. Defining G was well motivated by looking at the fact that OGE (where in this case G is intersection of those two circles) looked collinear; stronger thinking of $OG\cdot OE=OC\cdot OA=OB^2$ would even let us define it as the inverse, which made the chasing cyclicity easier.

Wow I still can't believe I solved this. Here's a sketch because there's no way I'm writing this nonsense up for real. I'll prove the converse problem. A phantom points argument plus some knowledge about the locus of $A$ that satisfies the condition being a circle should suffice to show the converse implies the original statement. Now if we have $\angle BXA+\angle DXC=180$, then this means that the quadrilateral formed by taking the perpendiculars at $A$, $B$, $C$, and $D$ to $XA$, $XB$, $XC$, $XD$ is cyclic. We can now rewrite the problem with respect to this new quadrilateral, and the length condition rewrites with LoS as $\frac{XA\cdot XB}{AB^2}=\frac{XC\cdot XD}{CD^2}$. Let $E=AB\cap CD$ and $F=AD \cap BC$. The angle condition stays the same actually when we rewrite it. Now angle chasing yields $XACF$ cyclic and $EDBX$ cyclic. Define $X'$ symmetrically. Let $M=(ADX)\cap (BCX)$. Then $M,X,F$ are collinear by radical axis. Angle chasing also yields $BDM$ is a line. Define $N$ symmetrically to $N$. More angle chasing yields $ABX'M$ and $CX'MD$ are cyclic, so $M=EX' \cap FX$, and $N=EX \cap FX'$. Angle chase to get $MXNX'$ cyclic. Now we can rewrite the length condition from earlier with a lot more LoS to get that it suffices to show $MA\cdot MC=MB\cdot MD$. More angle chasing yields that $MXNX'$ and $ABCD$ have a common Miquel point, say $H$. Yet more angle chasing implies we have the similar quadrilaterals $EBXD ~ EAX'C$ and $FAXC ~ FBX'D$. Using this and LoS allows us to rewrite our desired length condition as $AM \cdot CM = BM \cdot DM$. Now we also have the spiral similarities (by angle chasing) $HAM ~ HNC$ and symmetric variants. This allows us to rewrite our length condition as $BM\cdot DM \cdot HN = AN \cdot CN \cdot HM$. Extend $HN$ to intersect $(HAOC)$ at $L$, and let $HM$ intersect $(HBDO)$ at $K$. I now claim we have the spiral congruence $OMK \equiv OLN$. We can get our two equal lengths by angle chasing to get $\angle MHO = \angle NHO$ and we can get $\angle OKM=\angle ONL$ by straightforward angle chasing. This completes the proof.

It is known the desired statement is equivalent to $X$ having an isogonal conjugate in $ABCD$. Invert at $X$ with radius 1. Note we have the angle equalities and length relationships of the form \[\measuredangle A^* = \measuredangle D^*A^*X + \measuredangle X^*A^*B^* = \measuredangle XDA + \measuredangle ABX = \measuredangle XBC + \measuredangle ABX = \measuredangle B,\]\[A^*B^* \cdot C^*D^* = \frac{AB \cdot CD}{XA \cdot XB \cdot XC \cdot XD} = B^*C^* \cdot D^*A^*,\] which is enough to imply the similarity $A^*B^*C^*D^* \sim BCDA$. Notice that $X$ wrt $A^*B^*C^*D^*$ corresponds to the isogonal conjugate of of $X$ wrt $ABCD$, as desired. $\blacksquare$

Solved with CT17 and asdf334 Note that it suffices to show that $X$ has an isogonal conjugate. Consider an inversion around $X$ with radius $1$. Note that $$\angle C'A'B'=\angle XA'D'+\angle XA'B'=\angle XDA+\angle XBA=\angle XBC+\angle XBA=\angle CBA.$$ Furthermore, since $$A'B'=\frac{AB}{XA\cdot XB},$$we have $$A'B'\cdot C'D'=B'C'\cdot D'A'$$since $XA\cdot XB\cdot XC\cdot XD$ cancels on each side. Claim: $ABCD\sim A'B'C'D'$. We claim that if the angles are fixed and the product of opposite sides are equal, then the quadrilateral is unique up to scaling. Fix segment $AB$, and fix lines $AD$ and $BC$ as well since we are fixing the angles of the quadrilaterals. WLOG $AC$ and $BD$ intersect at a point that is closer to $D$ than $A$ and closer to $C$ than $B$. Then, we pick $C$ on line $BC$, and then go in a specific direction (determined by the angle) to hit line $AD$ at $D$. However, $CD$ decreases as $C$ and $D$ move closer to the intersection point, but $AD$ and $BC$ both increase, so by monotonicity there can only be at most one point that works, proving the claim. Thus, after inverting, we have the same diagram, but the vertices are shifted and, more importantly, the angle conditions switch orientations. For example, we have $\angle XBA=\angle XA'B'$ and $\angle XBC=\angle XB'C'=\angle XA'D'$. This means that if a spiral similarity takes $A'B'C'D'$ to $BCDA$, then the image of $X$ is the isogonal conjugate of $X$, so we are done.

First note that such a point $X$ is unique. This is true because $X$ is the intersection of circles $(A, C, \overline{AB} \cap \overline{CD})$ and $(B, D, \overline{BC} \cap \overline{AD})$(which is true by angle chasing), with $X$ being the intersection that lies in $ABCD$ Inverting at $X$ with arbitrary radius sends $ABCD$ to $A^\ast B^\ast C^\ast D^\ast$. Then we can angle chase to find that $\angle A^\ast B^\ast C^\ast = \angle A^\ast B^\ast X^\ast + \angle C^\ast B^\ast X^\ast = \angle XAB + \angle XCB = \angle XCD + \angle XCB = \angle C$. So it follows that $ABCD$ has the same angles(however in a different order) as $A^\ast B^\ast C^\ast D^\ast$. Then by the Inversion Distance formula we get that $A^\ast B^\ast \cdot C^\ast D^\ast = A^\ast D^\ast \cdot B^\ast C^\ast$ so $ABCD \sim D^\ast A^\ast B^\ast C^\ast$. Then the spiral similarity sending $A^\ast B^\ast C^\ast D^\ast \to ABCD$ maps $X$ to its isogonal conjugate, $X'$(wrt $ABCD$). This is enough as this implies that $\angle BXA + \angle DXC = 180^{\circ}$(we can show this by taking the ellipse inscribed into $ABCD$ with foci $X$ and $X'$ and angle chasing).

This is equivalent to showing $X$ has an isogonal conjugate. Let $AB\cap CD=E$ and $AD\cap BC=F$. Then, $X=(ACE)\cap(BDF)$. Define $Y=(ACF)\cap(BDE)$. We claim $X$ and $Y$ are isogonal conjugates. Let $\angle XAB=x$, $\angle XBC=y$, $\angle YAD=x'$, and $\angle YBA=y'$. We need to show $(x,y)=(x',y')$. By the Law of Sines, we have $$\frac{\sin(C-x+y)\sin(D-y)}{\sin(D+x-y)\sin(y)}=\frac{BC}{CD},$$so swapping $(A,C)$ and $(B,D)$ and multiplying gives $$\sin(A-x+y)\sin(C-x+y)\sin(D-y)\sin(B-y)=\sin(B+x-y)\sin(D+x-y)\sin(y)^2.$$Similarly, $$\sin(A-x+y)\sin(C-x+y)\sin(x)^2=\sin(B+x-y)\sin(D+x-y)\sin(A-x)\sin(C-x).$$The same equations hold for $x'$ and $y'$. Note that the two curves are polynomials in $x_1=e^{\pm xi}$ and $y_1=e^{\pm yi}$. When $A=B=C=D=\frac{\pi}2$, the equations reduce to $\cos(x-y)^2\cos(2y)=0$ and $\cos(x-y)^2\cos(2x)=0$, which is equivalent to $(x_1/y_1+y_1/x_1)^2(y_1^2+1/y_1^2)=0$, or $(x_1^2+y_1^2)^2(y_1^4+1)=0$. Now, consider all $ABCD$ which are sufficiently close to a square. Then, $x$ and $y$ must be close to $\frac{\pi}4$. However, since the curves $(x_1^2+y_1^2)^2(y_1^4+1)=0$ and $(x_1^2+y_1^2)^2(x_1^4+1)=0$ intersect at the solution corresponding to $x=y=\frac{\pi}4$ with multiplicity $1$, this means that there can only be one intersection close to $x=y=\frac{\pi}4$ for the new quadrilateral, which implies $(x,y)=(x',y')$. Now, we are done by moving points.

Troll problem. Note that by taking the Clawson-Schmidt conjugate of $X$, say $ Y$, it can be observed that the final condition is equivalent to $X$ having an isogonal conjugate. Then consider inversion about $X$ and note that the image of $ABCD$ is similar to $A^*B^*C^*D^*$, yet the relations involving $X$ swap and become $\angle XB^*A^* = \angle XD^*C^*$ and $\angle XC^*B^* = \angle XA^*D^*$ which in fact finishes, as then $X$ in $A^*B^*C^*D^*$ corresponds to the isogonal conjugate of $X$ in $ABCD$.

Let $E=\overline{AB}\cap\overline{CD}$ and let $F=\overline{AD}\cap\overline{BC}$. Note that $\angle{}XAB=\angle{}XCD$ is equivalent to $X$ lying on $(ACE)$, that $\angle{}XBC=\angle{}XDA$ is equivalent to $X$ lying on $(BDF)$, and that $\angle{}BXA+\angle{}DXC=180^\circ$ is equivalent to $X$ lying on the isoptic cubic $\omega$ of complete quadrilateral $ABCDEF$. Therefore, it suffices that $(ACE),(BDF),$ and $\omega$ share a common point other than the circle points $I$ and $J$. Lemma. If $A,B,C,D,E,$ and $F$ are points on a cubic $\gamma$ with group operation $+$ which has constant point $N$, then $A,B,C,D,E,$ and $F$ are coconic if and only if $$A+B+C+D+E+F=2N.$$ Proof. Let $\overline{AB}$ intersect $\gamma$ at $A,B,$ and $X$, let $\overline{CD}$ intersect $\gamma$ at $C,D,$ and $Y$, and let $\overline{EF}$ intersect $\gamma$ at $E,F,$ and $Z$. We see that $A,B,C,D,E,F,X,Y,$ and $Z$ are the $9$ intersections of the cubics $\gamma$ and $\overline{AB}\cup\overline{CD}\cup\overline{EF}$, so by Cayley-Bacharach we see that $A,B,C,D,E,$ and $F$ are coconic if and only if $X,Y,$ and $Z$ are collinear if and only if $X+Y+Z=N$ if and only if $(N-A-B)+(N-C-D)+(N-E-F)=N$ if and only if $A+B+C+D+E+F=2N$, proving the lemma. Now, define a group operation $+$ on $\omega$ with neutral point $A$. Let $(ACE)$ intersect $\omega$ at $A,C,E,I,J,$ and $X'$ and let $(BDF)$ intersect $\omega$ at $B,D,F,I,J,$ and $X''$. It suffices that $X'=X''$, or that $2N-(A+C+E+I+J)=2N-(B+D+F+I+J)$, or that $A+C+E=B+D+F$. Now, note that $A+C+E=C+E=F$ and that $B+D+F=D+F+B=Z+B$, where $Z$ is the intersection of the tangent to $\omega$ at $A$ with $\omega$ other than $A$ and $A$, so it suffices that $Z$ lies on $\overline{BD}$. Note that $\overline{AZ}$ and $\overline{AC}$ are isogonal in $\angle{}DAB$, so if $\overline{AZ}\cap\overline{BD}=Z_1$, then it suffices that $Z_1$ has an isogonal conjugate in $ABCDEF$, or that the line isogonal to $\overline{BD}$ in $\angle{}ABC$, the line isogonal to $\overline{BD}$ in $\angle{}CDA$, and $\overline{AC}$ concur. We claim that they concur at the inverse of $\overline{AC}\cap\overline{BD}$ in the $B$-Apollonius circle of $\triangle{}ABC$ which also goes through $D$ since $AB\cdot{}CD=BC\cdot{}DA$. It suffices to prove the following. Claim. Let $D$ and $E$ be point on $\overline{BC}$ in $\triangle{}ABC$. Then, we have that $\overline{AD}$ and $\overline{AE}$ are isogonal in $\angle{}CAB$ if and only if $D$ and $E$ are inverses in the $A$-Apollonius circle $\omega$ of $\triangle{}ABC$. Proof. Let $X$ and $Y$ be the intersections of the internal and external angle bisectors of $\angle{}CAB$ with $\overline{BC}$, which lie on $\omega$. We see that $D$ and $E$ are inverses in $\omega$ if and only if $(X,Y;D,E)=-1$ if and only if $\overline{AX}$ and $\overline{AY}$ bisect $\angle{}EAD$ if and only if $\overline{AD}$ and $\overline{AE}$ are isogonal in $\angle{}CAB$ since $\angle{}XAY=90^\circ$, proving the claim. Therefore, we are done.

Amazing geometry problem; after failing this problem multiple times, it’s time I’ve had my revenge and provide what seems to be a different solution to everyone’s. We can actually rearrange the length condition to become $\frac{AB}{BC}=\frac{AD}{DC}$, from which we immediately deduce that $B$ and $D$ lie on an Apollonian circle $\omega$ of segment $AC$, so its centre $O$ lies on $\overline{AC}$. Invert the whole diagram around $\omega$. Now notice that $A\to C$ so essentially $ABCD$ as a quadrilateral is fixed. Now what happens to the other conditions? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -81.5226977877729, xmax = 74.91450732568423, ymin = -43.43751150875084, ymax = 42.602951303648815; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-23.469828702700916,-7.0169746932748485)--(-11.370388619706965,21.45948530003356)--(19.7779378946041,-7.516606583702635)--(-5.688963149164394,-14.051545102967975)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw((-5.688963149164394,-14.051545102967975)--(-72.03288501521364,-6.455936435362037), linewidth(0.5) + ffvvqq); draw((-72.03288501521364,-6.455936435362037)--(-11.370388619706965,21.45948530003356), linewidth(0.5) + ffvvqq); draw((-23.469828702700916,-7.0169746932748485)--(-11.370388619706965,21.45948530003356), linewidth(0.5) + ccqqqq); draw((-11.370388619706965,21.45948530003356)--(19.7779378946041,-7.516606583702635), linewidth(0.5) + ccqqqq); draw((19.7779378946041,-7.516606583702635)--(-5.688963149164394,-14.051545102967975), linewidth(0.5) + ccqqqq); draw((-5.688963149164394,-14.051545102967975)--(-23.469828702700916,-7.0169746932748485), linewidth(0.5) + ccqqqq); draw((-72.03288501521364,-6.455936435362037)--(19.7779378946041,-7.516606583702635), linewidth(0.5) + ffvvqq); draw((-23.469828702700916,-7.0169746932748485)--(-29.000405053022774,-20.0333816591836), linewidth(0.5) + ffvvqq); draw((-29.000405053022774,-20.0333816591836)--(-5.688963149164394,-14.051545102967975), linewidth(0.5) + ffvvqq); draw(circle((6.482937237130552,-8.59569385748992), 13.338720771748374), linewidth(0.5) + qqwuqq); draw(circle((-14.973113953724551,6.1815451057808035), 15.696976983213569), linewidth(0.5) + qqwuqq); draw((-11.370388619706965,21.45948530003356)--(-5.688963149164394,-14.051545102967975), linewidth(0.5) + ffvvqq); draw(circle((-2.0370281251628652,-23.806769534672352), 27.22612993821805), linewidth(0.5) + qqwuqq); draw(circle((-22.11076317887137,1.5311259316187524), 22.638355788083025), linewidth(0.5) + qqwuqq); draw((-6.783585427773017,-7.209747174434826)--(0.45800392382950533,3.3047956449190803), linewidth(0.5) + ffvvqq); draw((0.45800392382950533,3.3047956449190803)--(-11.370388619706965,21.45948530003356), linewidth(0.5) + ffvvqq); draw((0.45800392382950533,3.3047956449190803)--(19.7779378946041,-7.516606583702635), linewidth(0.5) + ffvvqq); draw((-5.688963149164394,-14.051545102967975)--(0.45800392382950533,3.3047956449190803), linewidth(0.5) + ffvvqq); draw((-23.469828702700916,-7.0169746932748485)--(0.45800392382950533,3.3047956449190803), linewidth(0.5) + ffvvqq); draw((-29.000405053022774,-20.0333816591836)--(0.45800392382950533,3.3047956449190803), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((-23.469828702700916,-7.0169746932748485),dotstyle); label("$A$", (-22.980962436721363,-5.794809028325989), NE * labelscalefactor); dot((19.7779378946041,-7.516606583702635),dotstyle); label("$C$", (20.283702102469125,-6.283675294305533), NE * labelscalefactor); dot((-11.370388619706965,21.45948530003356),dotstyle); label("$B$", (-10.881522353727412,22.681650964982417), NE * labelscalefactor); dot((-5.688963149164394,-14.051545102967975),dotstyle); label("$D$", (-5.259560294962546,-12.883369885029369), NE * labelscalefactor); dot((-72.03288501521364,-6.455936435362037),dotstyle); label("$O$", (-71.50093933519204,-5.18372619585156), NE * labelscalefactor); dot((-29.000405053022774,-20.0333816591836),dotstyle); label("$Y$", (-28.48070792899134,-18.871981643278776), NE * labelscalefactor); dot((0.45800392382950533,3.3047956449190803),dotstyle); label("$X$", (0.9734845962767606,4.471382557244423), NE * labelscalefactor); dot((-6.783585427773017,-7.209747174434826),dotstyle); label("$Z$", (-6.237292826921654,-6.039242161315761), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $X’$ be the inverted image of $X$, and let $Y=\overline{AB}\cap\overline{CD}$, and $Z=\overline{BD}\cap\overline{AC}$. Then: \begin{align*} \measuredangle XAB&=\measuredangle XAO+\measuredangle OAB\\ &=-(\measuredangle CX’O+\measuredangle OBC)\\ \measuredangle XCD=\measuredangle XCO+\measuredangle OCD\\ &=-(\measuredangle AX’O+\measuredangle ODA)\\ \measuredangle AX’O-\measuredangle CX’O&=\measuredangle OBC+\measuredangle ADO\\ \measuredangle AX’C&=\measuredangle ADO+\measuredangle OBC\\ &=-\measuredangle DCA+\measuredangle BAO\\ &=-\measuredangle DCA+\measuredangle BAC\\ &=-\measuredangle YCA-\measuredangle CAY\\ &=\measuredangle AYC \end{align*}so $AX’CY$ cyclic. Similarly, $BX’DY$ cyclic. This in fact immediately pins $X$ down up to two possibilities: $X’=Y$, or $X’$ is the Miquel point of self-intersecting quadrilateral $ACDB$. Before dealing with the cases we convert the conclusion into its inverted form: \begin{align*} \measuredangle BXA&=\measuredangle CXD\\ \measuredangle BXO+\measuredangle OXA&=\measuredangle CXO+\measuredangle OXD\\ \measuredangle X’BO+\measuredangle OCX’&=\measuredangle X’AO+\measuredangle ODX’ \end{align*} Let’s deal with the second case as the first case is a lot less interesting. In fact the first case doesn’t happen because its inverted image lies OUTSIDE of the quadrilateral. Notice that then $X’BAZ$ and $X’CDZ$ are cyclic. Hence we have: \begin{align*} \measuredangle X’BO+\measuredangle OCX’&=\measuredangle X’BD+\measuredangle DBO+\measuredangle ACX’\\ &=\measuredangle X’BD+\measuredangle DBO+\measuredangle AYX’\\ &=\measuredangle X’BD+\measuredangle DBO+\measuredangle BYX’\\ &=\measuredangle X’BD+\measuredangle DBO+\measuredangle BDX’\\ &=\measuredangle BX’D+\measuredangle DBO \end{align*}In fact $\measuredangle X’AO+\measuredangle ODX’$ is also equal to $\measuredangle BX’D+\measuredangle DBO$ by similar logic plus the fact that $BO=OD$, so they are in fact equal and we are done.