My favorite problem of the contest First observe that, by $AB \perp CD$ and $AD \perp BC$, we have that $C$ is the orthocenter of $\triangle ABD$. Now, let $B'$ and $D'$ be the foot of altitudes onto the sides $DA$ and $AB$. Then, our main claim is that the mid-point $G$ of the segment $B'D'$ is that desired point of the problem. We first state two very well-known properties of the symmedians. $\textbf{Lemma 1.}$ The $X$-symmedian of $\triangle XYZ$ is the locus of all points $P$ such that $$\frac{d(P,XY)}{d(P,XZ)}=\frac{XY}{XZ}.$$ $\textbf{Lemma 2.}$ Let $Y'$ and $Z'$ be the feet of altitudes of vertices $Y$, $Z$, respectively of $\triangle XYZ$. Then, the line joining the vertex $X$ to the mid-point of segment $Y'Z'$ is the$X$-symmedian of triangle $XYZ$. Now, we get back to the original problem. Lemma 1 suggests that the point we are desiring for lies on certain symmedians of the triangle(s). Let $C'$ be the mid-point of segment $AC$. Let the $B$-symmedian of $\triangle ABC$ meet the line $B'D'$ at point $G$. Now, as $\angle BAC = \angle BDD' = \angle BB'D'$, and $\angle B'BD' = \angle ABC$, we have that $$\triangle B'BD' \sim \triangle ABC,$$yielding that \begin{align}\frac{B'D'}{AC}=\frac{BB'}{BA}.\end{align}Again, as $BG$ is the $B$-symmedian of $\triangle ABC$, we have $\angle B'BG = \angle ABC'$. But, now, $$\angle GB'B = \angle D'B'B = \angle BAC',$$implies that $$\triangle B'BG \sim \triangle BAC',$$which yields \begin{align}\frac{B'G}{AC'}=\frac{BB'}{AB}.\end{align}Now, using equations $(1)$ and $(2)$, we obtain \begin{align*} \frac{B'D'}{AC} & = \frac{B'G}{AC'} \\\implies \frac{AC'}{AC} & = \frac12 = \frac{B'G}{B'D'}, \end{align*}which therefore implies that $G$ is the mid-point of segment $B'D'$, and thus, by Lemma 2, we have that $AG$ is the $A$-symmedian of $\triangle ABD$. Similarly chasing the angles for $\triangle ADC$, we obtain that $DG$ is also the $D$-symmedian of $\triangle ADC$. For the demonstration of the problem, we use Lemma 1 repeatedly. As $G$ lies on $A$-symmedian of $\triangle ABD$, we have that $GG_a = ka$, $GG_d = kd$. Again, as $G$ lies on $B$-symmedian of $\triangle ABC$, we have $GG_b = kb$. Finally, as $G$ lies on $D$-symmedian of $\triangle ADC$, we have $GG_c = kc$, for some positive constant $k$ as desired. $\blacksquare$

I couldn't solve this sad @below i got two constructions, but they yielded similarity for only two, not all four

Same here. Unfortunately could not solve or get any partial observations for this.

[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); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.52, xmax = 17.28, ymin = -9.8, ymax = 10.54; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); /* draw figures */ draw((-3.48,7.66)--(5.92,-1.74), linewidth(2) + dbwrru); draw((5.92,-1.74)--(-7.48,-1.64), linewidth(2) + dbwrru); draw((-7.48,-1.64)--(-0.83,5.01), linewidth(2) + rvwvcq); draw((-3.48,7.66)--(-3.549621874477919,-1.6693311800412096), linewidth(2) + rvwvcq); /* dots and labels */ dot((-3.48,7.66),dotstyle); label("$A$", (-3.4,7.86), NE * labelscalefactor); dot((5.92,-1.74),dotstyle); label("$B$", (6,-1.54), NE * labelscalefactor); label("$ax+by+c=0$", (0.76,3.48), NE * labelscalefactor,dbwrru); dot((-7.48,-1.64),dotstyle); label("$C$", (-7.4,-1.44), NE * labelscalefactor); label("$y=0$", (-1.26,-2.5), NE * labelscalefactor,dbwrru); dot((-3.51984962406015,2.320150375939851),linewidth(4pt) + dotstyle); label("$D$", (-3.44,2.48), NE * labelscalefactor); dot((-0.83,5.01),linewidth(4pt) + dotstyle); label("$E$", (-0.74,5.16), NE * labelscalefactor); dot((-3.549621874477919,-1.6693311800412096),linewidth(4pt) + dotstyle); label("$F$", (-3.46,-1.5), NE * labelscalefactor); label("$bx-ay+\lambda =0$", (-7.4,1.86), NE * labelscalefactor,rvwvcq); label("$x=0$", (-5.16,4.56), NE * labelscalefactor,rvwvcq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Easy bash. Observe that the equation of the line $AD$ becomes $x=0$ since its the $y$-axis and equation of line $BC$ becomes $y=0$ since its $x$-axis. Now let equation of line $AB$ be $ax+by+c=0$ for some real $a,b,c$. Note that we can easily get that the equation of line $CD$ will be $bx-a+c\lambda=0$ for some real $\lambda$ by the fact that $CD \perp AB$ and so product of slopes of the two lines is $-1$. Now $A$ will be the intersection of the lines $ax+by+c=0$ and $x=0$ . Solving we get the coordinates of $A$ as $(0,\frac{-c}{b})$. Similiarly we get the coordinates of $B$ as $(\frac{-c}{a},0)$ since its the intersection of the lines $ax+by+c=0$ and $y=0$. Similiarly we get coordinates of $C$ as $(\frac{-\lambda}{b},0)$ since its the intersections of the lines $bx-ay+\lambda=0$ and $y=0$. And similiarly solving for $D$ we get that coordinates of $D$ are $(0,\frac{\lambda}{a})$. Now using the distance formula for coordinate geometry we have that $$AB=\sqrt{\frac{c^2}{a^2}+\frac{c^2}{b^2}}=\frac{c}{ab}\sqrt{a^2+b^2}$$$$BC= \frac{\lambda}{b}-\frac{c}{a}$$$$CD=\sqrt{\frac{\lambda^2}{a^2}+\frac{\lambda^2}{b^2}}=\frac{\lambda}{ab}\sqrt{a^2+b^2}$$$$DA=\frac{\lambda}{a}+\frac{c}{b}$$. Now let $P(m,n)$ be the desired point. Note that by perpendicular distance formula onto a line we have $$d(P,AD)=m$$$$d(P,BC)=n$$$$d(P,CD)=\frac{bm-an+\lambda}{\sqrt{a^2+b^2}}$$$$d(P,AB)=\frac{am+bn+c}{\sqrt{a^2+b^2}}$$. And as we known $$|\frac{d(P,AD)}{AD}|=|\frac{d(P,BC)}{BC}|=|\frac{d(P,CD)}{CD}|=|\frac{d(P,AB)}{AB}|=k$$for some real $k$. Inputting our values we get $\implies$ $$|\frac{m}{\frac{\lambda}{a}+\frac{c}{b}}|=|\frac{n}{\frac{\lambda}{b}-\frac{c}{a}}|=|\frac{\frac{bm-an+\lambda}{\sqrt{a^2+b^2}}}{\frac{\lambda}{ab}\sqrt{a^2+b^2}}|=|\frac{\frac{am+bn+c}{\sqrt{a^2+b^2}}}{\frac{c}{ab}\sqrt{a^2+b^2}}|=k$$or we get the equation $$|\frac{m}{b\lambda+ac}|=|\frac{n}{a\lambda-bc}|=|\frac{bm-an+\lambda}{(a^2+b^2)\lambda}|=|\frac{am+bn+c}{(a^2+b^2)c}|=\frac{k}{ab}$$Now just check that $$m=\frac{n(ac+b\lambda)}{bc-a\lambda}$$works. So for any $n$ we have an $m$ . Hence atleast one point exists. Done. (we have a condition $n \neq 0$ ) $\blacksquare$.

This is my favourite problem on the test (even when I did not submit my solutions ) And my solution is really short compared to some of the above solutions I have 2 solutions infact 1. Elegant synthetic solution 2. Barycentric (which I did to see whether I have the patience to bash this one) And the above bash is pretty neat as well

You are IMOTCer ok.

Are the official solutions out yet? Too lazy to check

No, ig, submissions ended today but ill check

Where is the thread for P6?

P6,8 yet to be made @below ofc yes

Just asking, can I use a sharygin problem in my blog?

Durga01 wrote: No, ig, submissions ended today but ill check Someone told me not to discuss the problems until official solutions are out. Where is this rule written?

Here is my solution Solution: First of all,we get that such a quadrilateral is a concave quadrilateral as $AB\perp CD$ and $AD\perp BC$ ; where $B$ is the orthocentre of $\triangle ACD$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.739cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.3, xmax = 15.12, ymin = -5.62, ymax = 6.3; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); draw((2.6,5.56)--(2.479372106402213,0.9359307454181842)--(-0.02,-0.72)--(6.88,-0.9)--cycle, linewidth(2.) + red); /* draw figures */ draw((2.6,5.56)--(2.479372106402213,0.9359307454181842), linewidth(2.) + red); draw((2.479372106402213,0.9359307454181842)--(-0.02,-0.72), linewidth(2.) + red); draw((-0.02,-0.72)--(6.88,-0.9), linewidth(2.) + red); draw((6.88,-0.9)--(2.6,5.56), linewidth(2.) + red); /* dots and labels */ dot((2.6,5.56),linewidth(3.pt) + ududff); label("$A$", (2.68,5.68), NE * labelscalefactor,ududff); dot((-0.02,-0.72),linewidth(3.pt) + ududff); label("$C$", (0.06,-0.6), NE * labelscalefactor,ududff); dot((6.88,-0.9),linewidth(3.pt) + ududff); label("$D$", (6.96,-0.78), NE * labelscalefactor,ududff); dot((2.479372106402213,0.9359307454181842),linewidth(3.pt) + uuuuuu); label("$B$", (2.56,1.06), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.2222222222222223) * currentpicture; /* end of picture */ [/asy][/asy] Lemma 1: In any $\triangle XYZ$ and a point $P$ in the plane , and $PD$ and $PE$ be perpendicular lines from $P$ on $YZ$ and $XZ$. If $\frac{PD}{YZ}=\frac{PE}{XZ}$ then $P$ lies on $Z$-Symmedian of $\triangle XYZ$. Proof: Let $ZP$ intersect $\odot(XYZ)$ at $W$. $\frac{YZ}{XZ}=\frac{PD}{PE}=\frac{\frac{PD}{PZ}}{\frac{PE}{PZ}}=\frac{sin\angle PZY}{sin\angle PZX}=\frac{sin\angle WZY}{sin\angle WZX}=\frac{sin\angle WXY}{sin\angle WYX}=\frac{WY}{WX}$ $\implies \frac{YZ}{XZ}=\frac{WY}{WX} \implies WXZY$ is harmonic quadrilateral $\implies P$ lies on $Z$-Symmedian of $\triangle XYZ$ . [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.71cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.3, xmax = 15.12, ymin = -5.62, ymax = 6.3; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen zzttqq = rgb(0.6,0.2,0.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); draw((2.6,5.56)--(-0.02,-0.72)--(6.88,-0.9)--cycle, linewidth(2.) + zzttqq); /* draw figures */ draw((2.6,5.56)--(-0.02,-0.72), linewidth(2.) + zzttqq); draw((-0.02,-0.72)--(6.88,-0.9), linewidth(2.) + zzttqq); draw((6.88,-0.9)--(2.6,5.56), linewidth(2.) + zzttqq); draw(shift((3.490313946798893,1.502034627290907)) * scale(4.154484551658685, 4.154484551658685)*unitcircle, linewidth(2.)); draw((xmin, -0.5152876662087452*xmin + 2.6451791435161667)--(xmax, -0.5152876662087452*xmax + 2.6451791435161667), linewidth(2.)); /* line */ draw((-0.43307261565394306,2.868336120935404)--(-0.02,-0.72), linewidth(2.)); draw((-0.43307261565394306,2.868336120935404)--(2.6,5.56), linewidth(2.)); draw((3.914477993223439,0.6280969139625681)--(3.8766584573872516,-0.8216519597579283), linewidth(2.)); draw((3.914477993223439,0.6280969139625681)--(5.271779112677701,1.5273614327341227), linewidth(2.)); /* dots and labels */ dot((2.6,5.56),linewidth(3.pt) + ududff); label("$X$", (2.36,5.66), NE * labelscalefactor,ududff); dot((-0.02,-0.72),linewidth(3.pt) + ududff); label("$Y$", (-0.64,-0.88), NE * labelscalefactor,ududff); dot((6.88,-0.9),linewidth(3.pt) + ududff); label("$Z$", (6.82,-1.4), NE * labelscalefactor,ududff); dot((-0.43307261565394306,2.868336120935404),linewidth(3.pt) + uuuuuu); label("$W$", (-0.68,3.34), NE * labelscalefactor,uuuuuu); dot((3.914477993223439,0.6280969139625681),linewidth(3.pt) + xdxdff); label("$P$", (3.44,0.44), NE * labelscalefactor,xdxdff); dot((3.8766584573872516,-0.8216519597579283),linewidth(3.pt) + uuuuuu); label("$D$", (3.96,-0.7), NE * labelscalefactor,uuuuuu); dot((5.271779112677701,1.5273614327341227),linewidth(3.pt) + uuuuuu); label("$E$", (5.36,1.64), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.2) * currentpicture; /* end of picture */ [/asy][/asy] Now coming back to our original problem , let $j,k$ be the tangents to $\odot (ABD)$ at $B,D$ respectively and let $l,m$ be the tangents to $\odot (BCD)$ at $B,D$ respectively . Let $CF$ and $AE$ be the $C$-Symmedian of $\triangle BCD$ and $A$-Symmedian of $\triangle ABD$ repectively ; with $j \cap k=E$ and $l \cap m=F$ .Let $AE \cap CF = P$. We claim that P is the required point (that is) Let $PW,PX,PY,PZ$ be perpendiculars from $P$ on $CD,AD,AB,BC$ respectively. Then, $\frac{PW}{CD}=\frac{PX}{AD}=\frac{PY}{AB}=\frac{PZ}{BC}$ Proof: As $P$ lies on $C$-Symmedian of $\triangle BCD$ and $A$-Symmedian of $\triangle ABD$, $\frac{PD}{BC}=\frac{PG}{BC}$ and $\frac{PE}{AD}=\frac{PF}{AB}$--(by Lemma 1). So if we prove that $\frac{PD}{BC}=\frac{PE}{AD}$ , then we are done . Let $n,o$ be the tangents to $\odot ACD$ through $A,C$ respectively. Let $n \cap o= G$. So , we need to prove $DG,AE,CF$ are concurrent . Let $\angle CAD=A, \angle ACD=C , \angle ADC=D$. $\angle EBD=\angle BAD=\angle BCD=\angle EDB= 90^\circ-D$ --- (as $B$ is the orthocentre of $\triangle ACD$). As $EB=ED ; FB=FD$, we get $\angle EBD=\angle FDB=90^\circ-D$. Thus we observe that $\angle EBD=\angle FDB=90^\circ-D$ and $\angle FBD=\angle EDB = 90^\circ-D$. $\implies BF\parallel ED$ and $BE\parallel FD$. We also see that $BFDE$ is a rhombus (as opposite sides are parallel and adjacent sides are equal). So $EF\perp BD$ . But we know that as $B$ is the orthocentre of $\triangle ACD , BD\perp AC$ $\implies EF\parallel AC$. (2) Now $\angle CBE=\angle CBD-\angle EBD=180^\circ-A-(90^\circ-D)=90^\circ+D-A$ . $\angle GCB=\angle GCA+ \angle ACB= D+90^\circ-A=\angle CBE$. $\implies GC\parallel BE\parallel FD$. Similarly , $GA\parallel BF\parallel ED$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(22.692cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -2.86, xmax = 34.96, ymin = -8.06, ymax = 9.56; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen zzttqq = rgb(0.6,0.2,0.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ffqqff = rgb(1.,0.,1.); draw((8.4,4.6)--(3.9,-3.94)--(15.3,-4.22)--cycle, linewidth(1.6) + zzttqq); draw((8.4,4.6)--(8.274296422487224,-0.5179313701630545)--(3.9,-3.94)--(15.3,-4.22)--cycle, linewidth(2.) + yqqqyq); /* draw figures */ draw((8.4,4.6)--(3.9,-3.94), linewidth(1.6) + zzttqq); draw((3.9,-3.94)--(15.3,-4.22), linewidth(1.6) + zzttqq); draw((15.3,-4.22)--(8.4,4.6), linewidth(1.6) + zzttqq); draw((8.4,4.6)--(8.274296422487224,-0.5179313701630545), linewidth(2.) + yqqqyq); draw((8.274296422487224,-0.5179313701630545)--(3.9,-3.94), linewidth(2.) + yqqqyq); draw((3.9,-3.94)--(15.3,-4.22), linewidth(2.) + yqqqyq); draw((15.3,-4.22)--(8.4,4.6), linewidth(2.) + yqqqyq); draw(shift((14.037148211243613,1.9010343149184723)) * scale(6.249948457769445, 6.249948457769445)*unitcircle, linewidth(2.) + red); draw(shift((9.537148211243611,-6.638965685081524)) * scale(6.249948457769441, 6.249948457769441)*unitcircle, linewidth(2.) + green); draw(shift((9.662851788756392,-1.5210343149184729)) * scale(6.249948457769446, 6.249948457769446)*unitcircle, linewidth(2.)); draw((xmin, -2.3823619426673113*xmin + 19.19443752911879)--(xmax, -2.3823619426673113*xmax + 19.19443752911879), linewidth(1.2) + red); /* line */ draw((xmin, 0.2063134633436879*xmin-7.376595989158423)--(xmax, 0.2063134633436879*xmax-7.376595989158423), linewidth(1.2) + red); /* line */ draw((xmin, 0.20631346334368797*xmin-2.225030121818682)--(xmax, 0.20631346334368797*xmax-2.225030121818682), linewidth(1.6) + green); /* line */ draw((xmin, -2.382361942667314*xmin + 32.2301377228099)--(xmax, -2.382361942667314*xmax + 32.2301377228099), linewidth(1.6) + green); /* line */ draw((8.274296422487224,-0.5179313701630545)--(15.3,-4.22), linewidth(2.)); draw((10.264335751241079,-5.258925331397442)--(13.309960671246143,0.5209939612343879), linewidth(2.)); draw((3.9,-3.94)--(13.309960671246143,0.5209939612343879), linewidth(2.) + yqqqyq); draw((8.4,4.6)--(10.264335751241079,-5.258925331397442), linewidth(2.) + yqqqyq); draw((xmin, -2.3823619426673117*xmin + 5.351211576402516)--(xmax, -2.3823619426673117*xmax + 5.351211576402516), linewidth(1.2) + ffqqff); /* line */ draw((xmin, 0.2063134633436884*xmin + 2.866966907913018)--(xmax, 0.2063134633436884*xmax + 2.866966907913018), linewidth(1.2) + ffqqff); /* line */ draw((0.9596586202816273,3.0649574014909464)--(15.3,-4.22), linewidth(2.)); /* dots and labels */ dot((8.4,4.6),linewidth(3.pt) + ududff); label("$A$", (7.94,4.92), NE * labelscalefactor,ududff); dot((3.9,-3.94),linewidth(3.pt) + ududff); label("$C$", (3.48,-4.12), NE * labelscalefactor,ududff); dot((15.3,-4.22),linewidth(3.pt) + ududff); label("$D$", (15.5,-4.56), NE * labelscalefactor,ududff); dot((8.274296422487224,-0.5179313701630545),linewidth(3.pt) + uuuuuu); label("$B$", (8.36,-0.4), NE * labelscalefactor,uuuuuu); dot((10.264335751241079,-5.258925331397442),linewidth(3.pt) + uuuuuu); label("$E$", (10.34,-5.14), NE * labelscalefactor,uuuuuu); dot((13.309960671246143,0.5209939612343879),linewidth(3.pt) + uuuuuu); label("$F$", (13.38,0.64), NE * labelscalefactor,uuuuuu); dot((0.9596586202816273,3.0649574014909464),linewidth(3.pt) + uuuuuu); label("$G$", (1.04,3.18), NE * labelscalefactor,uuuuuu); dot((9.511837782758386,-1.2795877692869366),linewidth(3.pt) + uuuuuu); label("$P$", (9.6,-1.16), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.6666666666666667) * currentpicture; /* end of picture */ [/asy][/asy] Now applying Desargues Theorem on $\triangle ACG$ and $\triangle EFD$ , As $AC\cap EF ; CG\cap FD$ and $AG\cap ED$ are points at infinity in their respective directions, (as all three pairs of lines are parallel) , we get that the 3 intersection points are collinear (on line at infinity). So , $AE,CF$ and $GD$ are concurrent .Thus there exists a unique point such that the distance from it to the sides of the quadrilateral are proportional to the lengths of the sides respectively . $\blacksquare$.

Attached my solution

Attachments:
P19.pdf (65kb)

Note that $C$ is the orthocenter of $\triangle ABD$. So relabel point $D\to C$ and $C\to H$ to make it more intuitive. As remarked by many posts above, it suffices to show that the following four lines are concurrent: $A$-symmedian of $\triangle ABC$, $B$-symmedian of $\triangle ABH$, $C$-symmedian of $\triangle ACH$, and $H$-symmedian of $\triangle BHC$. Let $E,F$ denote feet of altitudes from $B,C$. We claim that the concurrency point is the midpoint $M$ of $EF$. To see why, we note the following fact. Fact: Let $ABCD$ be a cyclic quadrilateral and $P=AC\cap BD$, then line connecting $P$ and midpoint of $AB$ is $P$-symmedian of $\triangle PCD$. Proof: Obvious by similar triangle. $\blacksquare$ Apply this fact to $BCEF$ with point $A$, $BCEF$ with point $H$, $AEFH$ with point $B$ and $AEFH$ with point $C$, we are done.

Pretty much similar to above posts:

Attachments:
Problem 19 v2.pdf (48kb)

Problem 19. The opposite sidelines of quadrilateral $ABCD$ are perpendicular. Prove that there exists a point such that the distances from it to the sidelines are proportional to the lengths of the corresponding sides. [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); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -18.35, xmax = 10.35, ymin = -10.87, ymax = 10.87; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); pen ffqqtt = rgb(1,0,0.2); /* draw figures */ draw((-5.27,7.41)--(-8.99,-3.37), linewidth(0.8)); draw((-8.99,-3.37)--(4.55,-3.69), linewidth(0.8)); draw((4.55,-3.69)--(-5.27,7.41), linewidth(0.8)); draw((-8.99,-3.37)--(-5.450771612405165,-0.23889884989357782), linewidth(0.8)); draw((-5.450771612405165,-0.23889884989357782)--(4.55,-3.69), linewidth(0.8)); draw((-5.27,7.41)--(-5.450771612405165,-0.23889884989357782), linewidth(0.8)); draw((-5.27,7.41)--(-4.600658669663905,1.8641452456175364), linewidth(1.2) + blue); draw((-5.27,7.41)--(-2.22,-3.53), linewidth(1.2) + blue); draw((-4.600658669663905,1.8641452456175364)--(-6.737970397648042,-3.4232237424484952), linewidth(0.8) + qqccqq); draw((-8.99,-3.37)--(-4.600658669663905,1.8641452456175364), linewidth(0.8)); draw((-8.99,-3.37)--(-5.360385806202583,3.5855505750532113), linewidth(0.8)); draw((-5.450771612405165,-0.23889884989357782)--(-1.5534451271703466,3.2090062028096584), linewidth(1.2) + linetype("2 2") + red); draw((-5.450771612405165,-0.23889884989357782)--(-7.647872212157467,0.5192842884253984), linewidth(0.8) + linetype("2 2") + red); draw((-5.450771612405165,-0.23889884989357782)--(-5.526705474448575,-3.4518503876053512), linewidth(0.8) + linetype("2 2") + red); draw((-5.450771612405165,-0.23889884989357782)--(-2.22,-3.53), linewidth(1.2) + qqccqq); draw((-7.647872212157467,0.5192842884253984)--(-1.5534451271703466,3.2090062028096584), linewidth(0.8) + ffqqtt); /* dots and labels */ dot((-5.27,7.41),dotstyle); label("$A$", (-5.19,7.61), NE * labelscalefactor); dot((-8.99,-3.37),dotstyle); label("$B$", (-8.91,-3.17), NE * labelscalefactor); dot((4.55,-3.69),dotstyle); label("$D$", (4.63,-3.49), NE * labelscalefactor); dot((-5.450771612405165,-0.23889884989357782),linewidth(4pt) + dotstyle); label("$C$", (-5.37,-0.07), NE * labelscalefactor); dot((-5.360385806202583,3.5855505750532113),linewidth(4pt) + dotstyle); label("$E$", (-5.29,3.75), NE * labelscalefactor); dot((-2.22,-3.53),linewidth(4pt) + dotstyle); label("$F$", (-2.13,-3.37), NE * labelscalefactor); dot((-4.600658669663905,1.8641452456175364),linewidth(4pt) + dotstyle); label("$G$", (-4.53,2.03), NE * labelscalefactor); dot((-6.737970397648042,-3.4232237424484952),linewidth(4pt) + dotstyle); label("$H$", (-6.65,-3.27), NE * labelscalefactor); dot((-1.5534451271703466,3.2090062028096584),linewidth(4pt) + dotstyle); label("$I$", (-1.47,3.37), NE * labelscalefactor); dot((-7.647872212157467,0.5192842884253984),linewidth(4pt) + dotstyle); label("$J$", (-7.57,0.67), NE * labelscalefactor); dot((-5.526705474448575,-3.4518503876053512),linewidth(4pt) + dotstyle); label("$K$", (-5.45,-3.29), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution. Notations: Let $CB\cap AD=I, DC\cap AB=J, AC\cap BD=K$. Let $E$ is the midpoint of $AC$ and $G$ be the midpoint of $IJ$. $\delta(X, \ell)$ denotes distance of $X$ from line $\ell$. Proof: Clearly, $C$ is the orthocenter of $\triangle ABD$ as $BC\perp AD, CD\perp AB$. Also, as $BJID$ is cyclic because $\angle BJD=\angle BID=90$ and so, $IJ$ is anti-parallel to $BC$. Thus, $C$-symmedian of $\triangle BCD$ and $A$-symmedian of $\triangle ABD$ bisects $IJ$. Thus, $CG, AG$ are $C$-symmedian of $\triangle BCD$ and $A$-symmedian of $\triangle ABD$ respectively. Now, we note that, $AJCI$ is cyclic as $\angle AJC+\angle AIC=90+90=180$ and so, $\angle JAC=\angle JIC\implies \triangle BCA\sim\triangle BJI$ and as $G, E$ are midpoints of correponding sides $AC,JI\implies \triangle BEA\sim\triangle BGI\implies \angle ABE=\angle GBI\implies BG$ is $B$-symmedian of $\triangle ABC$ as $E$ is midpoint of $AC$. Also, we know that in any triangle $PQR$, if $X\in P$-symmedian, then $$\frac{\delta(X,PQ)}{PQ}=\frac{\delta(X, PR)}{PR}$$and thus, as $G$ lies on $A$-symmedian of $\triangle ABD, D$-symmedian of $\triangle BCG$ and $B$-symmedian of $\triangle ABC$, we have, $$\frac{\delta(G,AB)}{AB}=\frac{\delta(G,AD)}{AD}\quad \frac{\delta(G,BC)}{BC}=\frac{\delta(G,CD)}{CD}\quad \frac{\delta(G,AB)}{AB}=\frac{\delta(G,BC)}{BC}$$and so, $$\frac{\delta(G,AB)}{AB}=\frac{\delta(G,BC)}{BC}=\frac{\delta(G,CD)}{CD}=\frac{\delta(G,AD)}{AD}$$and thus, $G$ is the desired point. References: The proof of: In $\triangle ABC$, $A$-symmedian line bisects any line anti-parallel to $BC$ bounded by $AB,AC$ can be found in \textbf{Theorem 9.3} in the book \texttt{Lemmas in Olympiad Geometry, XYZ Press}. In any triangle $PQR$, if $X\in P$-symmedian, then $$\frac{\delta(X,PQ)}{PQ}=\frac{\delta(X, PR)}{PR}$$can be found in \textbf{Theorem 9.4} in the book \texttt{Lemmas in Olympiad Geometry, XYZ Press}. \textbf{Feedback:} Nice problem for symmedians .

We know that $AD\perp BC, AB\perp CD$ so let $ABCD $ be a concave quadrilateral, let $C$ be the orthocentre of $ABD.$ Let $CD$ intersects $AB$ at $F,$ $BC$ intersects $AD$ at $E. $ Let $K$ be the midpoint of $EF.$ Let $M,N,P,Q $ be the projections of $K$ to $AD, CB, AB, CD$ respectively. Note, $\widehat {DEB}=\widehat{DFB}=90^{\circ}\implies DEFB$ is a cyclic quadrilateral. So $\widehat{BEF}=\widehat{BDF}.$ Note that, $MK\perp AD, EB\perp AD\implies MK\parallel EB. $ Thus, $\widehat{MKE}=\widehat{KEB}=\widehat{FDB},\widehat{DFB}=\widehat{KME}=90^{\circ}. $ So $\triangle KME\sim\triangle DFB,$ therefore $\frac {MK}{DF}=\frac {EK}{DB}.$ Note, $NK\perp AB, CF\perp AB\implies KN\parallel CF$ so $\widehat{NKF}=\widehat{EBD}=\widehat {KFC}, \widehat {KNF}=\widehat {BED}=90^{\circ}\implies \triangle KNF\sim\triangle BED.$ Therefore, $\frac{KN}{BE}=\frac{KF}{BD}\implies \frac {MK}{DF}=\frac {KN}{BE}\quad\text {because} KF=EK, \frac {KM}{KN}=\frac {DF}{BE}.$ Note, $\widehat{ADF}=\widehat {ABE}\implies \triangle ADF\sim\triangle AEB\implies \frac{KM}{KN}=\frac{DF}{BE}=\frac{AD}{AB} \hdots (\clubsuit)$ Notice that, $\widehat{KEP}=\widehat{BDF},\widehat {BFD}=\widehat{KPE}=90^{\circ}\implies\triangle KPE\sim\triangle BFD.$ Now, $\frac {KP}{BF}=\frac {KE}{DB},\frac{KN}{BE}=\frac{KF}{BD}, KE=KF. $ After that, $\frac {KP}{BF}=\frac {KN}{BE}\implies \frac{KP}{KN}=\frac{BF}{BE}.$ $\widehat{AEC}+\widehat{AFC}=180^{\circ}\implies AFCE$ is a cyclic quadrilateral. $\widehat {BEF}=\widehat{BAC}\implies\triangle BEF\sim\triangle BAC. $ So $\frac {KN}{KP}=\frac {EB}{BF}=\frac {AB}{AC}\hdots (\diamondsuit)$ $\widehat {KFQ}=\widehat{DBE}, \widehat{DEB}=\widehat{KQF}=90^{\circ}\implies \triangle DEB\sim\triangle KQF.$ Therefore, $\frac{KQ}{DE}=\frac {KF}{DB}=\frac{KP}{BF}=\frac {EK}{DB}\implies\frac {KP}{KQ}=\frac {BF}{DE}, $ and $KE=KF.$ $\widehat{FBC}=\widehat{EDC},\widehat{BFC}=\widehat {DEC}=90^{\circ}\implies \triangle EDC\sim\triangle FBC.$ $\frac {KP}{KQ}=\frac{BC}{CD}=\frac {BF}{DE}\hdots (\heartsuit)$ From $(\heartsuit)\implies \frac {MK}{DF}=\frac{EK}{DB}=\frac{KQ}{DE}=\frac {KF}{DB}, \frac {KM}{KQ}=\frac {DF}{DE}=\frac {AD}{CD}\hdots (\bullet)$ So, $KE=KF,$ $\widehat{DFE}=\widehat{DAC}\implies \triangle DFE\sim\triangle DAC.$ We get from $(\clubsuit),(\diamondsuit), (\heartsuit), (\bullet)$ $\frac {KM}{KN}=\frac{AD}{AB},\frac {NK}{KP}=\frac {AB}{CB},\frac{KP}{KQ}=\frac {CB}{CD},\frac {MK}{KQ}=\frac {DA}{DC}. $ Therefore, $KM: KN: KP: QK=DA: AB: CB: CD. $ $K$ is the point such that the distance between $K $ and $DA, AB, CB, DC$ are proportional to the lengths of $DA, AB, CB, DC. $

mathsworm wrote: Just asking, can I use a sharygin problem in my blog? I made a blog post on this problem. It has a solution using barycentric coordinates and a synthetic solution

I really liked this one.

Attachments:
P19_diagram3.pdf (2kb)
P19_diagram1.pdf (3kb)
P19.pdf (102kb)

Quote: $\textbf{Lemma 1.}$ The $X$-symmedian of $\triangle XYZ$ is the locus of all points $P$ such that $$\frac{d(P,XY)}{d(P,XZ)}=\frac{XY}{XZ}.$$ Many of you have used this lemma in your solution, however, this is not exactly correct (this doesn't affect the solution in most cases, but still...). The locus consists of a pair of straight lines, one is indeed the symmedian and the other is the tangent to $(XYZ)$ at $X$. However, the converse is true, that is, for any point on the symmedian, the said ratios are equal. (You can't assume signed distances, as the problem asks for distances only, which are always positive)