I found this figure is rectangle such that $\angle ABD=\angle CAB= \angle ACD=\angle CDB=30^{\circ}$ Then the answer is $\boxed{\sqrt{3}}$

hi first your given details i can't understand what is point F?

Anar24 wrote: hi first your given details i can't understand what is point F?

Hello,Anar.I think IstekOympiadTeam wanted to write that

Let $G=\overline{FE}\cap\overline{CD}$. Then $G$ is midpoint of $CD$, and since $E$ is midpoint of $CA$, $\overline{FE}=\overline{GE}\parallel \overline{AD} \parallel \overline{CB}$. This yields that $BCFE$ and $ADFE$ are parallelograms where diagonals $BF$ and $AF$ are also angle bisectors. Therefore the parallelograms must be rombi, implying $|BC|=|EB|=|EF|=|EA|=|AD|$; and since $E$ is midpoint of the diagonals, $|DE|=|EB|=|EA|=|EC|$. The only parallelogram $ABCD$ that fulfills this is the rectangle whose diagonals form $30^\circ$ and $60^\circ$ angles with the sides.

Let $K=AD\cap CF$ and $L=BC\cap DF$. Note that $ADLC$ and $BDKC$ are parallelograms, hence $DK=BC=AD=CL$, thus $DCLK$ is also a parallelogram. Furthermore, $F$ is the midpoint of $CK$ and $DL$. As $AF,BF$ are angle bisectors of $\angle EAD,\angle CEB$, respectively, we obtain that $AK=AC$ and $BL=BD$. Hence, $DE=\frac{BD}{2}=\frac{BL}{2}=BC=AD=\frac{AK}{2}=\frac{AC}{2}=CE$, thus $DEFC$ is a rhombus, furthermore $CD\perp EF$. Also, $\angle EDF=60^\circ$, therefore $$CD=2\cdot \cos{30^\circ}\cdot DF=\sqrt{3}\cdot DF=\sqrt{3}\cdot AD.$$We conclude that $\dfrac{AB}{AD}=\boxed{\sqrt{3}}$. [asy][asy] size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair A,B,C,D,E,F,K,L; A=(0,0);D=(0,1);B=(1.73205080757,0);C=B+D-A;E=extension(B,D,A,C);F=C+D-E;K=extension(C,F,A,D);L=extension(B,C,D,F); draw(A--B--C--D--cycle,deep);draw(E--D--F--C--cycle,light);draw(A--E,deep);draw(B--E,deep);draw(D--K--F--L--C,deep);draw(E--F,deep);draw(A--F,med+dashed);draw(B--F,med+dashed); dot("$A$",A,W); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,W); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); [/asy][/asy]