We have that $\triangle{ADX}$ is similar with $\triangle{BCX}$, thus there exists 2 points $M,N$ on the perpendicular bisector of $(AB)$ such that $\triangle{AMN}$ is similar with $\triangle{ADX}$, and $\triangle{BMN}$ is similar with $\triangle{BCX}$. So we have that $\frac{AD}{AM}=\frac{AX}{AN}$ and $\angle{DAM}=\angle{XAN}$, but $\triangle{ADM}$ beeing similar with $\triangle{AXN}$, we have $\frac{AD}{AX}=\frac{DM}{XN}$, and similar $\frac{BC}{BX}=\frac{CM}{XN}$, thus we have $CM=DM$, therefore $M$ lies on both perpendicular bisectors of $AB$ and $CD$, therefore $M \equiv Y$. Now it follows easy, because $\angle{AYB}=2\angle{AYN}=2\angle{ADX}$.

Tiks wrote: Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX = \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX = \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB = 2\cdot\measuredangle ADX$.

Does this solution work? I am really unsure as to whether it works because it seems like too simple a solution especially for an ISL #6. Does anyone think that this problem is a little similar to IMO 1975 #3? Solution Define point $Y'$ to be on the perpendicular bisector of $AB$ and such that $\angle{AY'B}=2\angle{ADX}$. Define point $Z$ to be such that $\angle{AZY'}=\angle{BZY'}=\angle{AXD}=\angle{BXC}$. Note that since $Z$ and $Y'$ are on the perpendicular bisector of $AB$, $AY'=BY'$ and $AZ=BZ$. By AA similarity, we have that $\triangle{AXD} \sim \triangle{AZY'}$ and $\triangle{BXC} \sim \triangle{BZY'}$. This yields by spiral similarity that $\triangle{AZX} \sim \triangle{AY'D}$ and $\triangle{BZX} \sim \triangle{BY'C}$. Therefore, \[\frac{ZX}{DY'}=\frac{AZ}{AY'}=\frac{BZ}{BY'}=\frac{ZX}{CY'}\] This yields that $DY'=CY'$ and therefore that $Y'=Y$ is the intersection of the perpendicular bisectors of $AB$ and $CD$. Therefore $\angle{AY'B}=\angle{AYB}=2\angle{ADX}$.

Let $M,N$ are midpoints of the sides $AB$ and $CD$ and $DX\cap AB=P(1)$, $AX\cap CD=Q(.)$. Then since $ \triangle AXD \sim \triangle BXC$ we get that $ \angle BXP=\angle CXQ $, $ \angle AXP=\angle DXQ $ and \[ \frac{AX}{BX}=\frac{DX}{CX}. \] So \[ \frac{AP}{BP}=\frac{DQ}{CQ} \] and the parallel lines pass through $M$ and $N$ to $DX$ and $AX$ respectively, intersect on $AD$. Analoguosly, the parallel lines pass through $M$ and $N$ to $CX$ and $BX$ respectively, intersect on $BC$. Hence $ \angle AYB=2\angle ADX$.

Dear mathuz, i dont understand how the result follows. (rest is ok) Could you explain a bit ?

My solution: Let circumcerntres of $\triangle DAX$ and $\triangle BCX$ be $O_1,O_2$. Let $Y$ be point such that $AY=BY$ and $\angle AYB=2\angle ADX$. Since $\angle AO_1X=2\angle ADX$ we have $\triangle O_1AX$ ~$\triangle AYB$ so $\frac{AO_1}{AY}=\frac{AX}{AB}$ and since $\angle O_1AY=\angle XAB$ we have that $\triangle AO_1Y$~$\triangle AXB$. Similary we have $\triangle BO_2Y$~$\triangle AXB$. Now we have $O_1Y=\frac{AY}{AB}BX$ and $O_2Y=\frac{BY}{AB}AX$ and $O_1X=\frac{AX}{AB}AY$ and $O_2X=\frac{BX}{AB}BY$. Since $\angle AO_1Y=\angle YO_2B$ and $\angle DAX=\angle CBX$ we have $\angle DO_1Y=\angle CO_2Y$. Now by Law of Cosine we have $DY^2=O_1D^2+O_1Y^2-2O_1D\cdot O_1Y\cos(DO_1Y)=AY^2(\frac{AX^2+BX^2-2AX\cdot BXcos(DO_1Y)}{AB^2})$ and $CY^2=O_2Y^2+O_2C^2-2O_2Y\cdot O_2Ccos(CO_2Y)=AY^2(\frac{AX^2+BX^2-2AX\cdot BXcos(DO_1Y)}{AB^2})$ which implies $DY=CY$ so $Y$ is the intersection of perpendicular bisectors of $AB,CD$.

Feels too convenient, so perhaps I missed something? Tiks wrote: Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX = \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX = \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB = 2\cdot\measuredangle ADX$. WLOG let rays $\overrightarrow{BA}, \overrightarrow{CD}$ intersect. Let $M, N$ be midpoints of $\overline{AB}, \overline{CD}$ respectively. Let $P$ be the spiral center $\overline{AB} \mapsto \overline{DC}$. Construct $E, F$ outside $ABCD$ with $\triangle AEB \sim \triangle DYC$ and $\triangle AYB \sim \triangle DFC$. Observe that $\triangle AEY \sim \triangle DYF$ with spiral center $P$ again. Further, we see $k=\tfrac{EM}{MY}=\tfrac{YN}{NF}$. Let $L$ be the point on $\overline{DA}$ with $\tfrac{AL}{LD}=k$. Suppose $X$ is the point inside $ABCD$ with $\triangle LMN \sim \triangle XBC$. Lemma. $\overline{XL} \perp \overline{DA}$ and $\triangle XAD \sim \triangle XBC$. (Proof) By linearity, $\triangle AEY \mapsto \triangle LMN \mapsto \triangle DYF$ under spiral similarity with pivot $P$. Apply $\triangle PDA \sim \triangle PNM \sim \triangle PYE$ hence $\angle PME=\angle PLA$. Combined with $\angle PMB=\angle PLX$ we obtain $\angle XLA=\angle BME=90^{\circ}$. Now reflect $X$ in $L$ to get $X'$. Again linearity gives $\triangle X'AD \sim \triangle XBC$ and we're done. $\blacksquare$ Finally, we see $\triangle XAD \sim \triangle LMN \sim \triangle DFY$ hence $\tfrac{1}{2}\angle AYB=\angle AYE=\angle ADX$ and we're done here.

Assume $AB,CD$ aren't parallel; otherwise the problem doesn't make sense. Let $X_1,X_2$ be the reflections of $X$ over $BC,AD$ and $X_3,X_4$ be the projections of $X$ onto $BC,AD$. Let $M,N$ be the midpoints of $AB,CD$. Clearly $X_2AXD, XBX_1C$ are directly similar kites, so they're also similar to their direct average $X_4MX_3N$. Since $AX_4:X_4D=BX_3:X_3C$ and $AB,CD$ aren't parallel, this implies $X_4,X_3$ are the only such points on $AD,BC$ which divide segments $AD,BC$ in the same ratio with $X_3X_4\perp MN$. Next construct $Y_1,Y_2$ with $DYC\sim AY_1B, DY_2C\sim AYB$. Once again $AY_1BY, DYCY_2$ are directly similar kites, so for any $r$ their weighted average $r(AY_1BY)+(1-r)(DYCY_2)$ is also a kite. Since $Y_1M:MY=YN:NY_2$ we can choose $r$ so that this resulting kite contains $M,N$; then the other two vertices of the kite must lie on $AD,BC$, divide those two segments in the same ratio, and create a segment perpendicular to $MN$. This implies these other two vertices are precisely $X_4,X_3$ by our earlier work, hence $MX_4NX_3\sim Y_1AYB$ and $\angle AYB =\angle X_2DX=2\angle ADX$; the other equality follows similarly.

Let us deal with non-trivial cases with $AB \not \parallel CD$. Let $\odot (ADX)$ and $\odot (BCX)$ meet at $Z$, and $\odot (ABZ)$ meets $\odot (CDZ)$ at $W$. We claim that $W$ coincides with $Y$. Simple angle chasing gives $\angle AYB=\angle AWB$, and $\angle CYD=\angle CWD$. Also line $ZX$ is the internal angle bisector of $\angle AZB, \angle CZD$, so it suffices to prove that $ZW \perp XZ$. Now invert the diagram with centre $Z$. Denote by $T '$ by the image of any point $T$. Note that $X'=A'D' \cap B'C'$, and $W'=A'B' \cap C'D'$, and $ZX'$ bisects both $\angle A'ZB'$ and $\angle C'ZD'$. Assume that line $ZX'$ meets $A'B', C'D'$ at $U, V$, and a line perpendicular to $ZX'$ passing $Z$ meets $A'B', C'D'$ at $P, Q$, respectively. Then, $$ X'(A'B', UP)=Z(A'B',UP) = -1=Z(C'D', VQ)=X(C'D', VQ),$$so $P$ and $Q$ coincides with $W'$. Thus $X'Z \perp WZ'$, which implies $W \equiv Y$. $\square$

First, a long phantom point argument: Fix $\triangle AXD$ first and let $O$ denote its circumcenter. Let $B$ be as in the original problem. Then we redefine $Y$ such that $\angle AYB = 2 \angle ADX = \angle AOX$ and $AY = BY$. We then redefine $C$ such that $\angle DYC = 2 \angle DAX = \angle DOX$ and $YC = YD$. We will prove $\triangle AXD \sim \triangle BXC$ (oppositely oriented). This will solve the problem. [asy][asy] pair A = dir(125); pair D = dir(275); pair X = dir(-5); pair O = origin; draw(unitcircle); pair Y = 1.2*dir(-20); pair B = Y+(A-Y)*X/A; pair C = Y+(D-Y)*X/D; filldraw(A--O--X--cycle, invisible, deepgreen); filldraw(A--B--Y--cycle, invisible, deepgreen); filldraw(D--O--X--cycle, invisible, blue); filldraw(D--C--Y--cycle, invisible, blue); filldraw(B--X--C--cycle, invisible, grey); draw(circumcircle(B, X, C), grey); dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$X$", X, dir(X)); dot("$O$", O, dir(-X)); dot("$Y$", Y, dir(Y)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); /* TSQ Source: A = dir 125 D = dir 275 X = dir -5 O = origin R-X unitcircle Y = 1.2*dir(-20) B = Y+(A-Y)*X/A C = Y+(D-Y)*X/D A--O--X--cycle 0.1 lightgreen / deepgreen A--B--Y--cycle 0.1 lightgreen / deepgreen D--O--X--cycle 0.1 lightblue / blue D--C--Y--cycle 0.1 lightblue / blue B--X--C--cycle 0.1 yellow / grey circumcircle B X C grey */ [/asy][/asy] Anyways, the main idea is that $\triangle AOX \sim \triangle AYB$ and $\triangle DOX \sim \triangle DYC$. We break symmetry now and set up as follows. Since we have a similarity $\triangle AOY \sim \triangle AXB$, there should be a single complex number $t \in {\mathbb C}$ such that \begin{align*} y-a &= t(0-a) \implies y=(1-t)a \\ b-a &= t(x-a) \implies b=t(x-a)+a. \end{align*}Next since $\frac{c-y}{d-y} = \frac{x-0}{d-0}$ we have \[ c = y + \frac{(d-y)x}{d} = \frac{d(x+y)-xy}{d}. \]Now, we simply calculate \begin{align*} \frac{c-b}{x-b} &= \frac{\frac{d(x+y)-xy}{d}-[t(x-a)+a]}{x-[t(x-a)+a]} \\ &= \frac{dx + (d-x)[(1-t)a] - dt(x-a)-da}{d(1-t)(x-a)} \\ &= \frac{(1-t)[(d-x)a+dx-da]}{d(1-t)(x-a)} \\ &= \frac{(1-t)x(d-a)}{(1-t)d(x-a)} = \frac{x(d-a)}{d(x-a)} \\ &= \frac{\frac1d-\frac1a}{\frac1x-\frac1a} = \overline{\left( \frac{d-a}{x-a} \right)} \end{align*}as desired.

hint : let $O$ be the circumcircle of $\triangle ADX$ then use rotation and similarity properties. also use this technique. you should began from verdict and arrive to assumptions.

Here's a pretty nice complex solution that nobody's posted yet. Let $P, Q$, both inside $ABCD$, be points on the perp. bisector of $AB$ such that $\triangle{APQ}\sim \triangle{ADX}$ (note there is only one choice for $P$ and $Q$). For our setup, set $(ABP)$ be the unit circle, let $p=1$, and let $a, q, x$ be the free variables; note $b=\bar{a}=\frac{1}{a}$, and $q$ is real. We desire to prove $PC=PD$, or $|c-1|=|d-1|$. Now, note $A$ is the spiral center of $PQ, DX$, by the spiral center formula we have $a=\frac{px-qd}{p+x-q-d}=\frac{x-qd}{1+x-q-d}$, which we can rearrange to get $d=\frac{a+ax-aq-x}{a-q}$. Thus, we have $d-1=\frac{(a-1)(x-q)}{a-q}$, so changing $a$ for $\frac{1}{a}$ in this expression, we get $c-1=\frac{(1-a)(x-q)}{1-aq}$. Thus, it suffices to prove that $|a-q|=|1-aq|$, or $(a-q)(\frac{1}{a}-q)=(1-aq)(1-\frac{q}{a})$. But multiplying both sides by $a$, we find the equation indeed holds, so we are done.

Noone has posted a short and neat sollutiion using complex weighted gliding lemma, so here it is . Complex Weighted Gliding Lemma: Let $A_1B_1C_1$ and $A_2B_2C_2$ be two similar triangles. Construct points $X$, $Y$, $Z$ such that triangles $A_1XA_2$, $B_1YB_2$ and $C_1ZC_2$ are similar. Then $\triangle XYZ\sim\triangle A_1B_1C_1$. Proof sketch: As $A_1B_1C_1\sim A_2B_2C_2$ there exist linear function $f$ of complex plane mapping $A_1B_1C_1\mapsto A_2B_2C_2$. Also consider identity function $i$ as a linear function of complex plane. Then to get to points $X$, $Y$, $Z$ you just take weighted average of these functions, but average of linear functions is linear, hence $XYZ\sim A_1B_1C_1$. $\square$ Problem Proof: Denote $X_1$ reflection of $X$ by $AD$ and $X_2$ reflection of $X$ by $BC$. Note that $X_1AX\sim XBX_2$, construct $Y'$ such that $DY'C\sim X_1AX$. Note that $X_1DX\sim XCX_2$, hence if we can complex glide them to $AY'B$, hence $AY'B\sim X_1DX$. Because triangles $AY'B$ and $DY'C$ are isosceles, we get $Y'=Y$ and from similarities we got the angle condition, that problem wanted. $\square$

Solved with nukelauncher.

Let $O$ be the cirumcenter of $\triangle ADX$. Redefine $Y$ such that $\triangle AOX\sim \triangle AYB$, hence $\angle AYB=\angle AOX=2\angle ADX$. We want to show $Y$ lies on the perpendicular bisector of $\overline{CD}$. Let $(ADX)$ be the unit circle, so $O=0$. We fix $A,X,D$ and $B$ as arbitrary points, and will calculate $Y$ and $C$ based on these points. WLOG by rotation set $x=1$, so $OX$ is the real axis. From the spiral similarity sending $\triangle AOX$ to $\triangle AYB$, we have \[ \frac{y-a}{0-a} = \frac{b-a}{x-a} \implies y=\frac{-a(b-a)}{x-a}+a=a\cdot \frac{1-b}{1-a}\]Let $\bullet'$ denote the reflection over real axis of $\bullet$. The angle conditions imply that there exists a spiral similarity sending $A'\mapsto B$ and $D'\mapsto C$. Hence \begin{align*} \frac{c-x}{\bar{d}-x} = \frac{b-x}{\bar{a}-x} \implies c&=\frac{(\bar d-x)(b-x)}{\bar a-x} +x \\ &=\frac{\left(\frac1d-1\right)(b-1)}{\frac1a-1} + 1 \\ &= \frac{b/d-1/d-b+1/a}{1/a-1} \\ &= \frac{ab-a-abd+d}{d(1-a)}. \end{align*}Since we wanted to show $Y$ is on the perpendicular bisector of $CD$, it suffices to prove that $y$ satisfies \begin{align*} |y-d|=|y-c|\iff (y-d)(\bar y-\bar d)=(y-c)(\bar y-\bar c). \end{align*}So if $Z=\tfrac{y-d}{y-c}$, we need to prove $Z\cdot \bar Z=1$. Magically, \begin{align*} Z=\frac{y-d}{y-c}&=\frac{a\cdot\frac{1-b}{1-a}-d}{a\cdot\frac{1-b}{1-a}-\frac{ab-a-abd+d}{d(1-a)}}\\ &=\frac{a(1-b)-d(1-a)}{a(1-b)-\frac{ab-a-abd+d}d}\\ &=\frac{a-ab-d+ad}{a+\frac{a-ab}d-1}\\ &=\frac{ad-abd-d^2+ad^2}{ad+a-ab-d}\\ &=d. \end{align*}Hence $Z\cdot\bar Z=d\cdot \bar d=1$. The end.

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label("$A$", (0.19436570899429592,0.4542795161555022), NE * labelscalefactor); dot((0.7567611090605788,-0.4171962836832381),dotstyle); label("$B$", (0.7682378133883488,-0.38878001865019324), NE * labelscalefactor); dot((0.18023545231029714,-0.43181690860373295),dotstyle); label("$A'$", (0.19140760536339874,-0.40357053680467914), NE * labelscalefactor); dot((-1.4525810536799912,-1.1462058127995842),linewidth(4pt) + dotstyle); label("$Y$", (-0.7889109221157496,-0.7860196712682731), NE * labelscalefactor); dot((-0.6889109221157496,-0.7860196712682731),linewidth(4pt) + dotstyle); label("$C$", (-1.6782748621203722,-1.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $A'$ be the reflection of $A$ in $DX$. Redefine $Y$ as the point such that $A$ is the center of spiral similarity which sends $DA'$ to $BY$, then $\angle AYB=\angle ADA'=2\angle ADX$ and $AY=YB$. Therefore, it suffices to show $YD=YC$. Setting up the complex plane with $d=-1$ and $x=0$, then $a'=\overline{a}$. Notice that from the spiral similarity formula we have $$a=\frac{db-a'y}{d+b-a'-y}$$Changing the subject we have $$y=\frac{ab+b-a-a\overline{a}}{a-\overline{a}}$$Now notice that $X$ is the center of spiral similarlity sending $DC$ to $A'B$. Therefore, $$0=db-a'c$$Hence $$c=\frac{-b}{\overline{a}}$$Now $$y-d=\frac{ab+b-a-a\overline{a}+a-\overline{a}}{a-\overline{a}}=\frac{(b-\overline{a})(a+1)}{a-\overline{a}}$$Meanwhile $$y-c=\frac{ab+b-a-a\overline{a}}{a-\overline{a}}+\frac{b}{\overline{a}}=\frac{a(\overline{a}+1)(b-\overline{a})}{\overline{a}(a-\overline{a})}$$It is easy to see that they have the same modulus, so we are done.

complex solution without phantom points [asy][asy] unitsize(2cm); pair A, B, C, D, X, Y; X=(0,0); A=(-1,2); B=(-2,-2); C=(3,-1); D=A*(2*foot(C/B,(0,0),(1,0))-C/B); Y=extension((A+B)/2,circumcenter(A,B,C),(C+D)/2,circumcenter(B,C,D)); draw(A--B--C--D--A--X--D--C--X--B); draw(A--Y--B--C--Y--D); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$X$", X, W); label("$Y$", Y, S); [/asy][/asy] Since $\angle ADX=\angle BCX$ and $\angle DAX=\angle CBX$, this means that $\triangle ADX\sim\triangle BCX$, so $\frac{d-x}{a-x}=\frac{\overline c-\overline x}{\overline b-\overline x}$. since $X$ is inside $ABCD$. Assume without loss of generality $x=0$. Then, we have $\overline bd=a\overline c$. Now, since $|y-a|=|y-b|$, we must have \begin{align*} (y-a)(\overline y-\overline a)&=(y-b)(\overline y-\overline b)\\ -a\overline y-y\overline a+|a|^2&=-b\overline y-y\overline b+|b|^2\\ y(\overline b-\overline a)+\overline y(b-a)&=|b|^2-|a|^2. \end{align*}Similarly, we have $$y(\overline d-\overline c)+\overline y(d-c)=|d|^2-|c|^2.$$Therefore, we have $$y=\frac{(|b|^2-|a|^2)(d-c)-(b-a)(|d|^2-|c|^2)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$We need to show that $\frac{(a-y)(d-x)^2}{(y-b)(a-d)^2}$ is real. We have $$y-b=\frac{\overline a(b-a)(d-c)-(b-a)(|d|^2-|c|^2)+(b-a)(b\overline d-b\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}=\frac{(b-a)(\overline ad-\overline ac-d\overline d+c\overline c+b\overline d-b\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$Similarly, we have $$y-a=\frac{(b-a)(\overline bd-\overline bc-d\overline d+c\overline c+a\overline d-a\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$Therefore, we have $\frac{a-y}{y-b}=-\frac{\overline bd-\overline bc-d\overline d+c\overline c+a\overline d-a\overline c}{\overline ad-\overline ac-d\overline d+c\overline c+b\overline d-b\overline c}$. Since $\overline bd=a\overline c$, we also have $b\overline d=\overline ac$. Therefore, this means that $$\frac{a-y}{y-b}=-\frac{-\overline bc-d\overline d+c\overline c+a\overline d}{\overline ad-d\overline d+c\overline c-b\overline c},$$so $$\frac{(a-y)(d-x)^2}{(y-b)(a-d)^2}=\frac{(\overline bc+d\overline d-c\overline c-a\overline d)d^2}{(\overline ad-d\overline d+c\overline c-b\overline c)(a-d)^2}.$$Therefore, it suffices to show that $$\frac{(\overline bc+d\overline d-c\overline c-a\overline d)d^2}{(\overline ad-d\overline d+c\overline c-b\overline c)(a-d)^2}=\frac{(b\overline c+d\overline d-c\overline c-\overline ad)\overline d^2}{(a\overline d-d\overline d+c\overline c-\overline bc)(\overline a-\overline d)^2}.$$This is equivalent to $$\frac{(\overline bc+d\overline d-c\overline c-a\overline d)^2}{(\overline ad-d\overline d+c\overline c-b\overline c)^2}=\frac{(a-d)^2\overline d^2}{d^2(\overline a-\overline d)^2}.$$We will show that $\frac{\overline bc+d\overline d-c\overline c-a\overline d}{\overline ad-d\overline d+c\overline c-b\overline c}=\frac{(a-d)\overline d}{d(\overline d-\overline a)}$. Adding one to each side gives $$\frac{\overline ad-a\overline d+\overline bc-b\overline c}{\overline ad-d\overline d+c\overline c-b\overline c}=\frac{a\overline d-\overline ad}{d(\overline d-\overline a)},$$which is equivalent to $$\frac{a\overline d-\overline ad+b\overline c-\overline bc}{d(\overline d-\overline a)+\overline c(b-c)}=\frac{a\overline d-\overline ad}{d(\overline d-\overline a)}.$$Therefore, it suffices to show $\frac{a\overline d-\overline ad}{d(\overline d-\overline a)}=\frac{b\overline c-\overline bc}{\overline c(b-c)}$. Since $d=\frac{a\overline c}{\overline b}$, this means that we have \begin{align*} \frac{a\overline d-\overline ad}{d(\overline d-\overline a)}&=\frac{a\frac{\overline ac}b-\overline a\frac{a\overline c}{\overline b}}{\frac{a\overline c}{\overline b}(\frac{\overline ac}b-\overline a)}\\ &=\frac{\frac cb-\frac{\overline c}{\overline b}}{\frac{\overline c}{\overline b}(\frac{c-b}b)}\\ &=\frac{\overline bc-b\overline c}{\overline c(c-b)}\\ &=\frac{b\overline c-\overline bc}{\overline c(b-c)}. \end{align*} Therefore, this means that $\angle AYB=2\angle ADX$.

Let $Y',Z$ be points on perpendicular bisector of $AB,$ such that $AY'Z\stackrel{+}{\sim} ADX,BY'Z\stackrel{+}{\sim} BCX.$ By spiral similarity $AXZ\stackrel{+}{\sim} ADY',BXZ\stackrel{+}{\sim} BCY',$ so $$\frac{|AY'|}{|Y'D|}=\frac{|AZ|}{|ZX|}=\frac{|BZ|}{|ZX|}=\frac{|BY'|}{|Y'C|}\implies |Y'C|=|Y'D|\implies Y'=Y.$$Finally $\measuredangle AYX=2\measuredangle AYZ=2\measuredangle ADX,$ as desired.

Note that $\triangle AXD\sim \triangle BXC$. There exists a unique spiral similarity that takes $DX$ onto line $EY$. Let $D\to P$ and $E\to Q$. We have that $\triangle BQP\cong \triangle AQP$ so $\triangle BQP \sim \triangle BXC$. We also have $\triangle BQX\sim \triangle BPC$ and $\triangle AQX\sim\triangle APD$. Thus, \[\frac{DP}{CP}=\frac{AP}{BP}\cdot \frac{QX}{AQ}\cdot \frac{BQ}{QX}=1\]so $CP=DP$, implying $P=Y$ and the result follows.

Weird solution: Let $O$ be the circumcenter of $\triangle AXD$, and denote by $K, Z$ the images of $D, O$ under the spiral similarity taking $AD$ to $AB$. Now $DK/AD=BX/AX=BC/AD$ so $DK=BC$. Now in directed angles, \[ \angle ZKD = \angle ZKA + \angle AKD = \angle ZKA + \angle ABX = \angle ZKA + \angle ABZ + \angle ZBX = \angle KAB + \angle ZBX = \angle XBC + \angle ZBX = \angle ZBC \], so \triangle ZBC \cong \triangle ZKD$, whence $Z = Y$. This finishes.

Redefine $Y$ to the clawson schmidt conjugate of $X$ and let $P=(AXD)\cap(BXC)\cap(AYB)\cap(CYD).$ Angle chasing gives $PX$ bisects both $\angle APB$ and $\angle CPD.$ Now invert at $P.$ After inversion if we define $Y',Y''$ to be the intersections of the line through $P$ perpendicular to $PX$ with $AB,CD$ then prism lemma and the angle bisector harmonic bundles give $Y'=Y''=Y.$ Thus $\angle XPY=90^\circ$ so $PY$ is a bisector of $\angle APB$ and $\angle CPD$ as well. Thus $AY=BY,CY=DY$ so $Y$ is the same point as given in the problem. Now $\measuredangle AYB=\measuredangle APB=2\measuredangle APX=2\measuredangle ADX.$

@above Nice, I got the exact same solution.

let me cook translating unicode from word into latex on aops is annoying. What if we take Clawson Schmidt Conjugate of $Y$!!! Let $X'$ be the Clawson-Schmidt Conjugate of $Y$. We will prove that $X'=X. $ Let $P=(AYB\cap CPD)$. All angles are directed in this solution. Lemma: $\angle APB$ and $\angle CPD$ share an angle bisector. Let $E$ be the antipode of $Y$ in $(AYB)$ and let $F$ be the antipode of $Y$ in $(CPD)$. By the definition of perpendicular bisectors, $E,F$ are respective midpoints of arcs $AB, CD$ of their respective circles, not containing $Y. $ This means, by the Incentre Excentre Lemma, that $PE$ bisects $\angle APB$ and $PF$ bisects $\angle CPD.$ Yet by Thales, as $EY$ and $FY$ are diameters by symmetry, $\angle EPY=90^\circ$, and $\angle YPF=90^\circ$. Thus, $E,P,F$ are collinear, hence $\angle APB$ and $\angle CPD$ share an angle bisector. Corollary: This angle bisector is the line through $E,P,F.$ Now consider an inversion at $P$ with radius $PY$. Let $K^*$ be the inverse of $K$. Note $Y^*=Y.$ By Clawson-Schmidt conjugation’s definition, $APYB,PYCD,APX' D,BPX' C$ are cyclic. Hence, \[A^*,Y,B^* \quad C^*,Y,D^* \quad A^*,X'^* ,D^* \quad B^*,X'^* ,C^*\]are collinear (within each triple). We thus have the complete quadrilateral $A^* B^* YC^* D^* X'^{*}$. Clearly, $E^*$ is the intersection of the angle bisector of $\angle APB$ with $A^* B^*$, and similarly $F^*$ is the intersection with $C^* D^*$. Recall that by Thales, $\angle EPY=90^\circ=\angle YPF$, and since $E^* PF^*$ is the angle bisector, it is well-known that the two conditions imply : \[(A^*,B^*,E^*,Y)=-1=(C^*,D^*,Y,F^* )\]By the Prism Lemma,$ E^*,F^*,X'^{*}$ are collinear looking at our complete quadrilateral. This means that $E,F,X'$ are collinear. We proceed by straightforward angle relations in cyclic quadrilaterals. \begin{align} \angle ADX'=\angle APE=\angle EPB=\angle X' CB\\ \angle X' AD=\angle X' PD=\angle CPX'=\angle CBX' \end{align}Hence $X'$ works. Claim: Only one $X$ exists satisfying the properties in the problem, which is $X'$. Let us consider the locus of points $X_1$ such $\angle DAX_1=\angle X_1 BC$, without the acute angle restraint. Construct the point $X_1'$ such that $X_1 BX_1' C$ is a parallelogram. Let $P=AD\cap BC. $ As $P,A,D$ and $P,B,C$ are collinear, \[\angle PAX_1'=180^\circ-\angle X_1 AD-\angle X_1' AX_1=180^\circ-\angle X_1 BX_1'-\angle CBX_1=\angle X_1' BP\]Let $\Phi$ be the function that takes the isogonal conjugate of an object with respect to $\triangle ABP.$ Then, we have \[\angle \Phi(X_1' )AB=\angle PAX_1'=\angle X_1' BP=\angle AB\Phi(X_1' )\]This implies that $\Phi(X_1' )$ lies on $\ell_1$, which we define as the perpendicular bisector of $AB$. Since isogonal conjugation is an involution, and obviously all of the before steps hold for the converse, for any $\Phi(X_1')$ on $\ell_1, \Phi(\Phi(X_1' ))=X_1'$ which satisfies the properties in the question. Hence, letting $\mathcal{L}_1$ be the desired locus, we have \[\Phi(\mathcal{L}_1 )=\ell_1\]Theorem: The isogonal conjugate of a line with respect to a triangle is a circum-conic. Let $\ell_1$ intersect $(ABP)$ at $M_1$ and $M_2$ such that $M_1$ is the midpoint of minor arc $AB$, and $M_2$ the midpoint of major arc $AB$. Claim: $\mathcal{L}_1$ is a rectangular hyperbola. Consider the isogonal conjugate of $M_1$. We have: \[\angle APM_1=\angle M_1 PB=\angle M_1 AB=\angle PA\Phi(M_1 )\]By the Incenter Excenter Lemma, thus $PM_1\parallel A\Phi(M_1 )$. Similarly, \[\angle M_1 PB=\angle APM_1=\angle ABM_1=\angle \Phi(M_1 )BP\]Hence $PM_1\parallel B\Phi(M_1 )$. Note that $P,\Phi(M_1 ),M_1$ are collinear as $PM_1$ is the angle bisector of $\angle BPA$. This means $\Phi(M_1 )$ is the point at infinity along $PM_1$, and this lies on $\mathcal{L}_1. $ Now, $P,\Phi(M_2 ),M_2$ are clearly collinear by Thales as $90^\circ+90^\circ=180^\circ. $ \[\angle M_2 AB+\angle M_1 PA=(180^\circ-\angle BPA)/2+(\angle BPA)/2=90^\circ\]Hence $PM_1\perp A\Phi(M_2 )$ which implies $P\Phi(M_2 )\parallel A\Phi(M_2 )$. Analogously, we get $B\Phi(M_2 )\parallel P\Phi(M_2 )$. This implies that $\Phi(M_2 )$ is the point at infinity along $PM_2$. Both $\Phi(M_1 )$ and $\Phi(M_2 )$ belong to $\mathcal{L}_1$ as $M_1,M_2$ lie on $l_1$. This means that $\mathcal{L}_1$ has two distinct points at infinity (Note the parallel lines were orthogonal). As we also know that $\mathcal{L}_1$ is a conic, $\mathcal{L}_1$ must be a hyperbola. Let $M$ be the center of the hyperbola. By definition, $M\Phi(M_1 )$ and $M\Phi(M_2 )$ are asymptotes of $\mathcal{L}_1$. But $M\Phi(M_1 )$ is parallel to $M_1 P,$ which is orthogonal to $PM_2$ by Thales as $M_1 M_2$ is a diameter of $(\triangle ABP)$. Yet $PM_2$ is parallel to $M\Phi(M_2 )$, hence $M\Phi(M_1 )\perp M\Phi(M_2 )$ Thus $\mathcal{L}_1$ is a rectangular hyperbola. Recall that \[\angle PAX_1'=\angle X_1' BP\quad \Leftrightarrow\quad \angle X_1' AD=\angle CBX_1'\]Hence $X_1'$ also lies on $\mathcal{L}_1$. Note that constructing $X_1'$ is equivalent to reflection over the midpoint of $AB$ by the properties of parallelograms. Pick a branch of the hyperbola. Reflect that branch over the midpoint of $AB$. From our previous result, this must be a component of \mathcal{L}_1. Assume for the sake of contradiction that this reflection intersects the initial branch at some point $K$, other than the midpoint of $AB$. Then $K$ is fixed upon reflection but not the center of reflection, hence lie on different half-planes: obvious contradiction. Else, the reflection is distinct from the initial branch, yet part of $\mathcal{L}_1$, hence it is precisely the other branch of the hyperbola. This implies that the center of $\mathcal{L}_1$, $M$, is indeed the midpoint of $AB$. Note that even when we have $K$ as the midpoint of $AB$, for $K$ to lie on $\mathcal{L}_1$, it must have multiplicity $2$, one count from each branch of $\mathcal{L}_1$. Hence if we consider multiplicity, the midpoint of $AB$ is still the center of reflection hence the center of $\mathcal{L}_1$. Hence $M$ is always the midpoint of $AB$. Consequently, the asymptotes of $\mathcal{L}_1$ are more specifically the lines respectively parallel to, and perpendicular to the angle bisector of $\angle BPA$ through $M$. Now consider the locus of points $X_2$ such that $\angle ADX_2=\angle X_2 CB$. From all our previous reasoning, say by relabeling: \[A\Leftrightarrow D,B\Leftrightarrow C,X_1\Leftrightarrow X_2,X_1'\Leftrightarrow X_2'\]And so on. We have an analogous result for the locus of $X_2$, which in our original diagram is a rectangular hyperbola $\mathcal{L}_2$ with center $N$, which is the midpoint of $CD$. Its asymptotes are respectively the lines parallel to, and perpendicular to, the angle bisector of $\angle BPA$ through $N$. The locus of points $X$ satisfying the constraints in the problem is simply the intersection of $\mathcal{L}_1$ and $\mathcal{L}_2$. Bézout’s Theorem: If two plane algebraic curves of degrees $d_1$ and $d_2$ have no component in common, they have $d_1 d_2$ intersection points, counted with their multiplicity, and including points at infinity and points with complex coordinates. The two rectangular hyperbolas each have degree $2$. Hence by Bézout’s Theorem, there are $4$ intersection points. First of all, the points at infinity along the angle bisector of $\angle BPA$ and the orthogonal line, both of which lie on both $\mathcal{L}_1$ and $\mathcal{L}_2$ contribute to two intersection points, and are distinct. Further, as $P,A,D$ and $P,B,C$ are collinear, $P$ also lies on $\mathcal{L}_1$ and $\mathcal{L}_2$. This gives another intersection point, which lies on a different plane to the points at infinity hence is distinct. Note that P, and points at infinity, are outside of $ABCD$. We already know our point $X'$ from before satisfies the problem’s conditions, and since we constructed this point it must exist. If this is in the exterior of our quadrilateral, then no other point $X$ satisfying the conditions of the problem, can exist in the interior of $ABCD$ as we already have counted $4$ intersection points, hence the problem is vacuously true. Else, it is in the interior of the quadrilateral. Then, there are exactly $1+1+1+1=4$ intersection points already, which is the exact number allowed by Bézout, hence $X'$ is the unique point inside $ABCD$ that works. In fact, this means $X'=X$ by uniqueness. Hence, returning to our problem, by cyclic quadrilaterals, \[\angle APE=\angle ADX\quad \quad \quad \quad \quad \angle EPB=\angle XCB\]Hence, \[\angle AYB=\angle APB=\angle ADX+\angle XCB=2\angle ADX\]Because $\angle ADX=\angle XCB$ from the problem, and since $APYB$ is cyclic. This concludes the problem.

idk what in the world above is but here's another cs sol