Let $K$ be the intersection point of $AC$ and $BF.$ Also, let $L$ be the intersection point of $FC$ and $BD.$ Because $\triangle ABC$ is isosceles and $BF$ is the bisector of $\angle B,$ we see that $BF$ is a perpendicular bisector of $AC.$ Thus, $FD=AF=AB=AC.$ But $AD\parallel BC,$ and therefore $ABCF$ and $FBCD$ are parallelograms. This means that $L$ is the midpoint of $FC$ and of $BD,$ and thus $FE$ and $BL$ are equal medians of the isosceles $\triangle BCF.$ Finally, because $BD=2\cdot BL$ and $BL=EF,$ we get $BD/EF=2.$

Obviously $ABCF$ is a rhombus, $AD=2AF$ and, easily, $\Delta ABD\sim\Delta CEF$, their similitude ratio being $2$, hence the conclusion. Best regards, sunken rock

ANSWER IS 2

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BarisKoyuncu wrote: ANSWER IS 2 Where did you find the official s solution??

Let point $X \in AB$ such that $AX=BX$, obviously we have $AX=BX=EB=EC$. Because $BF$ is the bisector of $\angle XBE$ and $\triangle BXE$ is isosceles, $BXFE$ is a kite. We then have $$EF=XF=\frac{BD}{2} \rightarrow \frac{BD}{EF}=2$$Q.E.D