soroush.MG wrote: Suppose that there exists a point $P$ in $ABCD$ such that: $\angle APB > \angle APD , \angle DPC > \angle ABC$. It should be$\angle APB > \angle ADC$ and $\angle DPC > \angle ABC$ instead.

PART1. let w a circle with diameter AD and L is a line that is tangent to w at Q and is perpendicular to AB at m and to CD at N and let O the midpoint of AD WLOG:$AB <CD$ now let A' the perpendicular of A to CD we know it lie on w AMNA' is rectangle and AA' is a chord of w and we have OP is perpendicular to MN and MN is paralel to AA' now OQ is perpendicular to AA' at last we can see that OQ is perpendicular to the MN at midpoint of it at last we understand that Q is midpoint of MN and $MA=MA' $ it means that CQ is angel bisector # we now $ OQ=1/2 (AM+DN) OQ=1/2 (AD)$ PART2. let B' evidence to M and C' evidence to N its easy to show that the circumcircle of$ AQB' $and circumcircle of$ DQC'$ are tangent to each other let that line L2 and in # we know that $ADC =AQB' and ABC=DQC'$ $BC||B'C' $it means B and C are in $B'C'$ or out of it if BC there in B'C' its easy to show that the we can show that the Rainbow worthy of AB whit angel ADC is up of L2 and the Rainbow worthy of $CD $whit angel $ABC $is down of L2 and its possible now we have$ B' $is among of $AB$ and $C' $is among of $DC $ it means$ AB > AB' $ $DC> DC'$ $AB+CD>AB'+DC'=AD+BC $

Too trivial !! My proof: Let $\Omega$ be a circle tangent ton $AD$ and $BC$ at $A$ and $B$ ; Let $\omega$ be a circle tangent ton $AD$ and $BC$ at $D$ and $C$; Let $M,N$ be the midpoints of $BC$ and $AD$ respectively. Since $\angle ASB=\angle ADC\Longrightarrow P$ must be inside the circle $\Omega$. Similarly since $\angle DSC=\angle ABC\Longrightarrow P$ must be inside $\omega$. Hence $\omega$ and $\Omega$ must intersect each other at two distinct points $S,T$. easy to see that $ST$ passes throw $M,N$. Thus if $MT=SN=x,ST=y$ (see figure) then by AM-GM inquality: $$BC=2MB=2\sqrt{x(x+y)}<2x+y=MN=\frac{AB+CD}{2}\Longrightarrow AD+BC<AB+CD$$ Q.E.D [asy][asy] import graph; size(12.42475239102794cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -5.724817209911015, xmax = 35.69993518111692, ymin = -13.45539462448323, ymax = 25.611128038591136; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); /* draw figures */ draw(circle((10.077897977003042,9.461099706682875), 4.944425615875747)); draw(circle((10.39364081793557,-0.6539477331910236), 8.529668362410135)); draw((9.64245549762135,23.41081056401747)--(5.401758015313216,11.067667392656867)); draw((5.401758015313216,11.067667392656867)--(2.3267941934540017,2.117555091954764)); draw((2.3267941934540017,2.117555091954764)--(18.271924907828524,2.6152849247334546)); draw((18.271924907828524,2.6152849247334546)--(14.644733170136668,11.3561883562411)); draw((14.644733170136668,11.3561883562411)--(9.64245549762135,23.41081056401747)); draw((5.401758015313216,11.067667392656867)--(14.644733170136668,11.3561883562411)); draw((16.458329038982598,6.985736640487278)--(14.318748831830447,6.918949298458), red); draw((15.472614180225039,6.754642813958068)--(15.460120026196282,7.15490196266506), red); draw((15.316957844616763,6.749783976280218)--(15.304463690588006,7.150043124987211), red); draw((14.318748831830447,6.918949298458)--(6.003856311535769,6.659398584335082), blue); draw((6.003856311535769,6.659398584335082)--(3.8642761043836087,6.592611242305816), red); draw((5.018141452778205,6.428304757805879)--(5.005647298749449,6.828563906512872), red); draw((4.8624851171699275,6.42344592012803)--(4.849990963141171,6.8237050688350225), red); label("$x$",(15.054301336394083,8.34710663395235),SE*labelscalefactor); label("$x$",(4.286535408758465,7.946652529205574),SE*labelscalefactor); label("$y$",(9.892892875213374,6.522815712328148),SE*labelscalefactor); /* dots and labels */ dot((5.401758015313216,11.067667392656867),linewidth(3.pt)); label("$A$", (4.242040508231045,11.105790466652362), NE * labelscalefactor); dot((2.3267941934540017,2.117555091954764),linewidth(3.pt)); label("$D$", (0.860428068147132,1.9843358585313544), NE * labelscalefactor); dot((18.271924907828524,2.6152849247334546),linewidth(3.pt)); label("$C$", (18.969852582807036,2.696254266970067), NE * labelscalefactor); dot((14.644733170136668,11.3561883562411),linewidth(3.pt)); label("$B$", (15.143291137448923,11.773213974563655), NE * labelscalefactor); dot((9.64245549762135,23.41081056401747),linewidth(3.pt) + uuuuuu); dot((6.003856311535769,6.659398584335082),linewidth(3.pt) + uuuuuu); label("$S$", (6.1553212309101015,7.3237239218217), NE * labelscalefactor,uuuuuu); dot((14.318748831830447,6.918949298458),linewidth(3.pt) + uuuuuu); label("$T$", (13.630464519516647,7.635188225513637), NE * labelscalefactor,uuuuuu); dot((3.8642761043836087,6.592611242305816),linewidth(3.pt) + uuuuuu); label("$N$", (2.729213890298768,6.700795314437826), NE * labelscalefactor,uuuuuu); dot((16.458329038982598,6.985736640487278),linewidth(3.pt) + uuuuuu); label("$M$", (16.92308715854572,7.234734120766861), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

The idea of problem is the same as G8-2006 ISL . https://artofproblemsolving.com/community/c6h155712p875030

bgn wrote: soroush.MG wrote: Suppose that there exists a point $P$ in $ABCD$ such that: $\angle APB > \angle APD , \angle DPC > \angle ABC$. It should be$\angle APB > \angle ADC$ and $\angle DPC > \angle ABC$ instead. Sorry! Edited.

Here's an INTERESTING way to prove the inequality: Target is to prove $P$ exists while this inequality is correct in the isosceles trapezoid. As Andria said, $P$ has to be inside both circles so we just have to prove that the inequality is correct while the two circles intersect at two points. WLOG suppose $AB<DC$. The idea is to consider that the circle which includes points $D$ and $C$ is fixed and move the other circle to see what happens when the circles distance changes. (When the circle gets far from the other $AD=BC$ gets bigger and $AB$ gets smaller while $DC$ doesn't change) We can guess from the inequality that when the circles are tangent, we have $AD+BC=AB+CD$ Having regard that $2MN=AB+DC$ , we can easily prove our guess So we know only when the two circles are close enough to intersect the inequality is correct