سلام من از راه حلم عكس دارم ولي چطور بايد بذارمش تو اين جا؟

See here too Iran second round 2016 day 2-p6 http://artofproblemsolving.com/community/q1h1235174p6270026

Finally i found Here's my solution Sorry! A little in farsi

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My solution: Let $x,y$ be the tangent of $(ATD)$ and $(ATB)$ pass through $A$. We have $\angle XAC=\angle ATD+\angle DAC=\angle ABC$$ \Longrightarrow $ $x$ is the tangent of $(ABC)$. Similarly, we have $y$ is the tangent of $(ADC)$. From $\angle BCA=\angle ACD$ $ \Longrightarrow $ $\angle ATB=\angle ATD$ $ \Longrightarrow $ $TA$ is the bisector of $\angle DTB$. Let $TD$ cut $CB$ at $L$. We will define the problem again: Suppose $A$ is the incenter of $\triangle DAC$ .$B$ is a point on $LC$ and $T$ is a point on the ray $LD$ sucth that $TA$ is the bisector of $\angle DTB$.$ \Longrightarrow $ $\angle TAB=\angle DAC=90^o+\frac{\angle DLC}{2}$ $ \Longrightarrow $ $\angle ABC=\angle DAC+\angle ATB$ and $\angle ADC=\angle BAC+\angle ATB$$ \Longrightarrow $ done.

$\angle ABC$ + $\angle CAB$ = $\angle ADC$ + $\angle CAD$ $ \Longrightarrow $ $\angle CAD$ + $\angle ATD$ +$\angle CAB$ = $\angle ATB$ + $\angle CAB$ + $\angle CAD$ $ \Longrightarrow $ $\angle ATD$ = $\angle ATB$ it means $C$ & $T$ move on a hyperbola($B$ & $D$ are focal points). Let $C$ & $T$ go to $ \infty$ you see their hyperbola is the same $ \Longrightarrow $ $\angle CAD$ = $\angle TAB$. Good luck in your olympiad

It' easy to prove <ATB=<ATD. K on CD that ABCK cyclic quadrilateral . Intersection BK AND AD is X . L on BC that CDAL cyclic quadrilateral . Intersection DL and AB is Y . It's easy to prove XTBA and YTDA cyclic quadrilateral .Intersection BK and DL is M . We have<BXT=<BAT=<YDT .===》》》XDMT cyclic quadrilateral .===》》》》<DTM=<MXD=<ATB=<ATD .===》》》 A ,T,M lies on line . it's easy to prove AMKD cyclic quadrilateral .===》》》<MAD=<MKC=<BAC . ===》》》<BAT=<DAC.

Awesome Geometry. Let $\angle BCA = \angle ACD = a$, $\angle BAC = y$, $\angle DAC = x$, $\angle BTA = b$ and $ATD = t$. Let $M,N$ be points on extension of $CB,CD$. $\angle 180 - a - x - b = y$ and $\angle 180 - a - y - t = x$ so $b = t$. Let $S$ be on $AC$ such that $\angle SBC = x$. Now we have $SBC$ and $DAC$ have spiral similarity and also $ABC$ and $DSC$. we have $\angle ABS = \angle ABC - x = t$ and $ADS = \angle ADC - y = b$. Let $AT$ meet $BS$ at $K$. Note that for proving $\angle BAT = x$ we need to prove $BAK$ is tangent to $BC$ or $\angle BKA = \angle ABM$. $\angle BKA = b - \angle TBK = t - \angle TBK = \angle ABT$. Let $BT$ meet $DS$ at $Q$. $\angle ADQ = b = \angle ATB \implies ATQD$ is cyclic so $\angle AQD = ATD = t = \angle ABS \implies ABSQ$ is cyclic. $\angle BKA = \angle ABT = \angle ABQ = \angle ASQ = \angle 180 - \angle DSC= \angle 180 - \angle ABC = \angle MBA$ so $\angle BKA = \angle ABM$ so we're Done.

Let $\odot (ABT),\odot (ADT)$ intersect $AC$ again at $P,Q,$ and $AD,AB$ at $X,Y$ respectively. From $\measuredangle AXP=\measuredangle ATP=\measuredangle ATB+\measuredangle BAP=\measuredangle ADC$ we obtain $PX\parallel CD$ and analogously $PY\parallel CB.$ Therefore $$\measuredangle YBX=\measuredangle ATX=\measuredangle ACD=\measuredangle BCA=\measuredangle YTA=\measuredangle YDX,$$which yields $AXYT\stackrel{-}{\sim} ABDC,$ and the conclusion follows.