Got by Brianchon that the diagonals and $WZ,XY$ concur, where $W,X,Y,Z$ are the tangency points of the incircle with the quad. Then I tried setting the incircle as the unit circle and get $wz+xy=0$ as well as get the coordinates of $A,B,C,D$ and then using the concylic equation but did not help

The only problem I liked this year. NJOY wrote: The diagonals of bicentric quadrilateral $ABCD$ meet at point $L$. Given are three segments equal to $AL$, $BL$, $CL$. Restore the quadrilateral using a compass and a ruler. Construction: Construct collinear points $A,L,C$. Draw a circle $\omega$ with center $L$ and radius $BL$. Now construct a point $X$ such that $AX = AL + BL$ and $CX = CL + BL$ which we can by drawing circles with center $A$ and $C$ with respective radiuses. Now, draw the internal and external bisectors of $\angle AXC$ and let it meet $AC$ at $Y,Z$. Draw the circle $\odot(XYZ)$ and construct one of its intersections with $\omega$ is $B$. Now draw $\odot(ABC)$ and extend $BL$ to meet it at $D$ and we are done. Proof of construction: Note that from the construction $B$ is a point such that $BL$ is given and $\frac{AB}{BC} = \frac{AL + BL}{CL + BL}$ as $\odot(XYZ)$ is the Apollonius circle. Therefore we need to show that $\frac{AB}{BC} = \frac{AL + BL}{CL + BL}$ and we will be done. Let the incircle of $ABCD$ meet $AB,BC,CD,DA$ at $Q,R,S,P$ respectively. We know, $PR,QS,AC,BD$ meet at $L$. Note that \[90^{\circ} = \frac{180^{\circ} - \angle QAP + 180^{\circ} - \angle SCR}{2} =\angle APQ + \angle CRS = \angle PSQ + \angle RPS.\]Therefore $PR\perp QS$. From $\triangle BLR$ and $\triangle QLB$, \[\tan \angle BLR =\frac{\sin \angle BLR}{\cos \angle BLR} = \frac{\sin \angle LRB}{\sin \angle BQL} = \frac{\sin \angle PSR}{\sin \angle QPS} = \frac{PR}{QS}.\] Similarly $\tan \angle CLS = \frac{QS}{PR}$, so $\tan \angle RLC = \frac{PR}{QS} = \tan \angle BLR$, so $LR$ bisects $\angle BLC$. This gives us $\frac{BL}{BL+CL} = \frac{BR}{BR+RC} \implies BR = \frac{BL\cdot BC}{BL+CL}$ and similarly $BQ = \frac{BL\cdot AB}{BL+AL}$. This gives us \[\frac{BL\cdot BC}{BL+CL} =\frac{BL\cdot AB}{BL+AL} \implies \frac{BC}{AB} = \frac{BL+CL}{BL+AL}\]and we are done. $\hfill{\blacksquare}$

Solution : Let $W,X,Y,Z$ be the touchpoints of the incircle of the bicentric quadrilateral $ABCD$ with sides $AB,BC,CD,DA$ respectively . Let $AB\cap CD=E$ and $AD\cap BC=F$ . So, $\triangle EWY$ and $\triangle FXZ$ are isosceles . So , $\angle AWL=\angle DYL=\theta$.Also,as $ABCD$ is cyclic quadrilateral , $\angle WAL=\angle BAL=\angle BAC=\angle BDC=\angle LDC=\angle LDY$.So $\triangle AWL\sim \triangle DYL$ . Similarly , $\triangle BWL\sim \triangle CYL \implies \frac{BL}{BW}=\frac{CL}{CY}$. So , $\frac{AL}{AW}=\frac{DL}{DY}$.Applying sine rule in $\triangle AWL$ and $\triangle CYL$, we get , $\frac{AL}{CL}=\frac{AW}{CY} \implies \frac{AL}{AW}=\frac{CL}{CY}=\frac{DL}{DY}=\frac{BL}{BW}=\frac{1}{k}$ for some $k$. So , $AW=k\cdot AL , BW=BX=k\cdot BL , CY=CX=k\cdot CL$. Now,$\frac{AB}{BC}=\frac{AW+WB}{BX+XC}=\frac{AW+BW}{BW+CY}=\frac{k\cdot(AL+BL)}{k\cdot(CL+BL)}=\frac{AL+BL}{BL+CL}=p$ . We know that locus of point $B$ such that $\frac{AB}{BC}=p$ is a circle $\omega$ whose diameter lies on $AC$ --- (1) (EGMO Theorem 9.19-- Appolonian Circles) Point $B$ also lies on the circle $\Omega$ with centre $L$ and radius $BL$. So , $B=\omega\cap \Omega$. Construction of $\omega$ : Let $G$ be a point on $AC$ such that $\frac{AG}{GC}=p$. So, $G$ lies on $\omega$.That is divide $AC$ in the ratio $\frac{AL+BL}{BL+CL}=p$ (dividing it in the usual way of dividing a line segment in a particular ratio). Now let $H$ be a point on line $AC$ such that $(H,G;A,C)=-1$,so even $H$ lies on $\omega$. To construct this point , consider any point $P$ not on $AC$. Let $Q$ be any point different from $P$ and $Y$ on $PY$. Let $AQ\cap CP=R$ and $CQ\cap AP=S$ . Let $RS\cap BC=H$.We claim that $H$ is the required point. Proof: By Lemma 9.11 - Cevians induces harmonic bundles lemma of EGMO, $(H,G;A,C)=-1$. From (1) , $GH$ is the diameter of $\omega$.Thus , we draw $\omega$. Now , $B=\omega\cap \Omega$. Thus , we get point $B$. Now we can draw the circumcircle of $\triangle ABC$ which is also the circumcircle of $ABCD$. As $L=AC\cap BD , BL\cap \odot (ABC)= D$. 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Attached my solution

Attachments:
P21.pdf (69kb)

This was my solution

Attachments:
Problem 21.pdf (40kb)

Construct the bicentric quadrilateral $ABCD$ having $3$ segments equal to $A\ell, B\ell, C\ell.$ Stage 1- To determine the value of $\ell D, $ Construct $2$ points $A’, C’$ on a plane. Let $\ell’$ is on $A’C’$ and $\ell A=\ell’A’, \ell C=\ell’C’. $ We have $A’C’ = AC.$ Let any $B’$ on that plane and $\ell B =\ell’B’.$ Let $B’\ell’$ intersects $\odot(A’B’C’)$ at the second point is $D’,$ so we have $\ell D = \ell’D’.\quad$ Stage 2- Construct $\ell $ between $A, C$ that satisfies the lengths given from the beginning. $\quad $ Stage 3- Construct the segment that is equal to $\frac{(\ell A+\ell C)(\ell B+\ell D)}{2\sqrt{\ell B\cdot\ell D}}.$ $\text {WLOG},$ we suppose $\ell B>\ell D.$ Construct $X,Y, Z$ such that $X, Y, Z$ are collinear and $XY = \ell D, XZ = \ell B.$ Construct the circle $(\omega)$ with $XZ$ being the diameter. Let $(\omega)$ intersects the line through $Y$ and perpendicular to $XZ$ at $T.$ Let $W$ be the reflection of $T$ to $X.$ We have $WT = 2XT = 2\sqrt {\ell B\cdot\ell D}; YZ=\ell B-\ell D.$ Let $N$ be a point on segment $MP$ such that $MN = \ell A, PN =\ell C\implies MP = MN + PN = \ell A+\ell C.\quad$ Construct $\triangle X_1X_4X_6$ in which $\widehat{X_1X_6X_4}$ is right-angled. Let $X_2,X_3$ respectively be a point on $X_1X_6,X_1X_4$ such that $X_2X_3\parallel X_4X_6.$ By $\text {Thale's theorem}$ $X_2X_3=X_4X_6\,\, \frac{X_1X_3}{X_1X_6}=\frac{(\ell A+\ell C)+\left|\ell B+\ell D\right|}{2\sqrt{\ell B\cdot\ell D}}.$ We have constructed the segment $X_2X_3=\frac{(\ell A+\ell C)+\left|\ell B+\ell D\right|}{2\sqrt{\ell B\cdot\ell D}}.\quad$ Stage 4- Construct a point $K$ such that $AK\perp AC$ and $X_2X_3=\frac{(\ell A+\ell C)+\left|\ell B+\ell D\right|}{2\sqrt{\ell B\cdot\ell D}}.\quad$ Stage 5- Construct a circle with diameter $KC(\omega_2)$ and the circle with center $\ell,$ radius $\ell D(\omega_3).$ $(\omega_2)$ doesn't intersect $(\omega_3).$ Claim- We cannot construct the bicentric quadrilateral $ABCD.\quad$ Proof- Let $AB = a, BC=b,CD=c, DA = d.\,\triangle LAB\sim\triangle LDC\implies\tfrac ac=\frac{\ell B}{\ell C}.$ Similarly, $\tfrac bd=\frac{\ell B}{\ell A}. $ Since $ABCD$ is a bicentric quadrilateral, we have $$AD+BC = AB + CD\implies a+c=b +d\implies a-d=b-c\implies a^2+d^2-2ad=b^2+c^2-2bc\implies a^2+d^2-b^2-c^2=2(ad-bc)\,(\star).$$Based on the $\text{Cosine theorem},BD^2=a^2+d^2-2ad\cos\widehat {BAD}=b^2+c^2+2bc\cos\widehat{BAD}.$ Thus $a^2+d^2-b^2-c^2=2\cos\widehat{BAD}(ad+bc)\,(\bullet).$ From $(\star) $ and $(\bullet)$ $$ad-bc=ad+bc\cos\widehat{BAD}\implies\cos\widehat{BAD}=\left(\frac{ad-bc}{cd}\right):\left(\frac{ad+bc}{cd}\right)=\left(\tfrac ac-\tfrac bd\right):\left(\tfrac ac+\tfrac bd\right)=\frac {\ell A-\ell C}{\ell A+\ell C}.$$Similarly, $\cos\widehat{ABC}=\frac{\ell B-\ell D}{\ell B+\ell D}. $ Note, $\sin^2\widehat{ABC}=\frac{4\ell B\cdot\ell D}{(\ell B+\ell D)^2}.$ $\text{WLOG},$ we suppose that $\ell B >\ell D$ (other cases can be solved similarly). Thus, $0^{\circ}<\widehat{ABC}< 90^{\circ}\implies\sin\widehat{ABC}=\frac{2\sqrt{\ell B\cdot\ell D}}{\ell B+\ell D}\implies\tan\widehat{ABC}=\frac{2\sqrt{\ell B\cdot\ell D}}{\ell B-\ell D}.$ A line through $C$ and perpendicular to $C$ at $C$ intersects the circumcircle of quadrilateral $ABCD$ at $K_1.\,\, A$ is the diameter of the circumcircle of $ABCD\,\text { and}\,CK_1= AC.$ Note, $\cot\widehat{AK_1C}=AC,\cot\widehat{ABC}=\frac {(\ell A+\ell C)(\ell B-\ell D)}{2\sqrt{\ell B\cdot\ell D}}.$ Thus, is identical to K or is the reflection of K to line AC. If $K_1$ is identical to $K\implies (\omega)$ is the circumcircle of $ABCD.\,\, D$ is the intersection of $(\ell;\ell D)$ and $(\omega)\implies\mathrm {contradiction}.$ Suppose that $K_1$ is the reflection of to line $AC.$ Let $B’, D’$ respectively be the reflection of $B, D$ to $AC.$ Because $ABCD$ is a bicentric quadrilateral. Notice that $(\omega_2)$ is the circumcircle of $ABCD.\,\, D’$ is the intersection of $\ell; \ell D$ and $(\omega_2).$ $(\omega_2) $ intersects $(\omega_3)$ at $D_1,D_2. $ Proof- Let $\widehat{AKC}=\alpha(0^{\circ}<\alpha<90^{\circ}).$ Suppose that $D_1$ lies on the half-plane of line $AC$ that does not contain $K.$ Let $AB_1= x, B_1C=y, CD_1= z, D_1A = t.$ Notice that, $z^2+t^2+2zt\cos\alpha=x^2+y^2-2xy\cos\alpha\implies x^2+y^2-z^2-t^2=2cos\alpha(xy+zt)\,(\square).$ Also, $\ell D_1=\ell D_2=\ell D,\overline{\ell B_1}=\overline {\ell B_2}=\frac{\overline{\ell A}\cdot\overline{\ell C}}{\overline{\ell D}}=\overline{\ell B}.$ Note, $\triangle\ell AD_1\sim\triangle\ell B_1C\implies\tfrac ty=\frac{\ell D}{\ell C},\tfrac zx=\frac{\ell D}{\ell A}.$ Note, $0 <\alpha <90^{\circ}\implies\cos\alpha>0^{\circ}\,,\sin\alpha>0^{\circ}.$ Note, $$\tan\alpha=\frac{AC}{AK}=\frac{2\sqrt{\ell B\cdot\ell D}}{\left|\ell B-\ell D\right|}\implies\cos\alpha=\sqrt{\frac{1}{1+\tan\alpha^2}}=\frac{\ell B-\ell D}{\ell B+\ell D}.$$Note, $$\frac{xy-zt}{xy+zt}=\frac{\tfrac xy-\tfrac ty}{\tfrac xz+\tfrac ty}=\frac{\frac{\ell A}{\ell D}-\frac {\ell D}{\ell C}}{\frac {\ell A}{\ell C}-\frac{\ell D}{\ell C}}=\frac{-\ell D^2+\overline{\ell A\cdot\ell C}}{\ell D^2+\overline{\ell A\cdot\ell C}}=\frac{\ell B-\ell D}{\ell B+\ell D}\implies\frac{xy-zt}{xy+zt}=\frac{\ell B-\ell D}{\ell B+\ell D}=\cos\alpha. $$ From $(\square)$ $$x^2+y^2-z^2-t^2=2\cos\alpha(xy+zt)=2(xy-zt)\implies (x-y)^2=(z-t)^2\implies |B_1A-B_1C|=|D_1A-D_1C|\implies |B_2A-B_2C|=|D_2A-D_2C|\,(\diamondsuit).$$ Let $O$ be the center of $(\omega).$ If $B_1A-B_1C=D_1A-D_1C, B_1A+CD_1= AD_1+CB_1\implies AB_1CD_1$ is a bicentric quadrilateral. If $B_1A-B_1C=D_1C-D_1A.$ $D_2$ is the reflection of $D_1$ to $O\ell,B_2$ is the reflection of $B_1$ through $O\ell.$ Therefore, $B_2D_2=B_1D_1.$ Let $AB_2= x',B_2C=y',CD_2= z',D_2A = t'.$ Suppose that $B_2A-B_2C\ne D_2A-D_2C\, (*)$ Then from $(\diamondsuit)$ $$B_2A-B_2C=-D_2A+D_2C\implies x'-z'=y'-t'$$Note, $$x-z=y-t\implies\frac {x'-z'}{x-z}=\frac{y'-t'}{y-t}\,(\Gamma). $$$\triangle \ell AB_2\sim\triangle\ell D_2C\implies \frac {x'}{z'}=\frac {\ell B_1}{\ell C}=\frac{\ell B}{\ell C}.$ Note, $\triangle \ell AB_1\sim\triangle\ell D_1C\implies\frac{x'}{z'}=\frac{\ell B_1}{\ell C}=\frac{\ell B}{\ell C}.$ $\triangle\ell AB_1\sim\triangle\ell D_1C\implies\tfrac xz=\frac{\ell B_1}{\ell C}=\frac{\ell B}{\ell C}.$ After that, $$\frac{x'}{z'}=\tfrac xz\implies\frac {x'}{x}=\frac {z'}{z}\implies\frac{x'-z'}{x-z}\,(\Delta)\implies\frac{y'}{y}=\frac {t'}{t}=\frac {y'-t'}{y-t}\,(\nabla).$$ From above $$\frac {x'}{x}=\frac {y'}{y}\implies\frac {x'}{y'}=\frac{x}{y'}\implies\frac{AB_2}{CB_2}=\frac{AB_1}{CB_1}\implies B_1\equiv B_2\implies\text{contradiction}.$$ Note, $(*) $ is false $\implies B_2A-B_2C=D_2A-D_2C\implies B_2A+D_2C=D_2A+B_2C\implies AB_2CD_2$ is a bicentric quadrilateral $\cdot\quad $ $(\omega)$ touches $(\omega)$ at $D.$ Let $O$ be the center of $(\omega),$ $B$ be the second intersection of $\ell D$ to $(\omega).$ Consider $AB = t, BC = y, CD = z, DA = t\implies |x-y|=|z-t|.$ If $x-y = z-t\implies x+t = y + z\implies B, O,\ell, D$ are collinear. Note, $\widehat{DAB}=\widehat{DCB}=90^{\circ}.$ Note, $x^2+t^2=y^2+z^2\implies xt=\frac {(x+t)^2-x^2-t^2}{2}=\frac{(y+z)^2-y^2-z^2}{2}=yz.$ Therefore, $\{x,t\}=\{y,z\}.$ Since, $x-y=z-t\implies x = z, y = t\implies\frac{\ell B}{\ell C}=\tfrac xz=1. $ Cause $\triangle\ell BA\sim\triangle \ell CD\implies\ell B=\ell C.$ Similarly, we get that $\ell A =\ell B =\ell C =\ell D\implies AK=\frac {(\ell A+\ell C)(\ell B-\ell D)}{2\sqrt{\ell B\cdot\ell D}}=0.\, A\equiv K.$ Hence, $(\omega)$ is the circle with diameter $AC,$ center $\ell$ and $(\omega)$ is the circle with center $\ell,$ radius $\ell D (=\ell A=\ell C)\implies(\omega_2)$ is identical to $(\omega_3)\implies\text{contradiction}\implies x-y=t-z,$ or $x +z =y+t\implies ABCD$ is the bicentric quadrilateral. $(\omega)$ is identical to $(\omega).$ Proof- $\ell $ is the circumcircle of $(\omega_3),\ell $ is also the circumcircle of $(\omega_2)\implies\ell A =\ell B =\ell C.$ Construct the lines $d_1,d_2\perp\ell.$ Construct the distinct points $B, D$ such that $\ell B =\ell D(B, D\,\text { lies on}\, d_2)$ and $2$ distinct points $A,C$ such that $$\ell A =\ell C (A, C\,\text {lies on}\, d_2 ).AB = AL\sqrt {2} = CL\sqrt {2}=CD,BC=AD.$$Also, $AB = BC\implies AB = BC= CD = DA.$ Since $BD\perp AC, ABCD$ is a square$\implies ABCD$ is a bicentric quadrilateral.