About The Triangle 
As you can see in image $A$, $D$, $E$, $F$ are  linear. $BD$ $\bot$ $DC$, $A$$(BEC) = 8 cm^2$ and $A$$(ABC) = 18 cm^2$. so what $cm^2$ $A$$(BDCE)$ is?
I couldn't solve this problem, thanks for everyone...
 A: Edit
Just after posting the stuff below, I realized that there is a simpler explanation as to why your problem is underspecified. And this one is more fundamental. It cannot be overcome by assuming that the triangle is isoceles and that $\overline{AF}\perp\overline{BC}$. 
There is nothing in the problem that forces a specific size to $BC$. Imagine moving $B$ and $C$ closer together. $A$ and $E$ both must move farther away from $\overline{BC}$ in order to meet the area requirements on $\triangle ABC$ and $\triangle BEC$. But $D$ must move closer to $\overline{BC}$ in order to maintain a right-angle. Thus moving $B$ and $C$ closer together shrinks the quadrilateral. It cannot have a fixed area determinable by the information given.
Original post
Here is fuller account of my comment, explaining why additional assumptions are required to calculate a specific area for $\square BDCE$.
Nothing in your problem description specifically requires $AB = AC$ or $\overline{BC} \perp \overline{AF}$ as is seen in your diagram. In this example, I am assuming that we are given a right triangle $\triangle ABC$ with the right angle at $B$. We are given that the area is $18$ (since I'm dealing with this in the abstract, there is no reason to use specific units of measure, such as centimeters). 



*

*Draw a circle having $\overline{BC}$ as diameter. We are only interested in the part inside the triangle. But since $\angle ABC$ is a right-angle, $\overline{AB}$ is tangent to the circle, so the arc from $B$ to the intersection with $\overline{AC}$ is inside.

*Choose a point $D$ on this arc and draw $\triangle BDC$. Since this triangle is inscribed in the circle and the side $\overline{BC}$ is a diameter, $\angle BDC$ is a right angle. 

*Draw $\overline{AD}$ and extend it until it intersects $\overline{BC}$ in the point $F$.

*Pick the point $E$ on $\overline{AF}$ such that the area of $\triangle BEC$ is 8. (To be definitive on how this can be done, divide $\overline{AF}$ into $9$ equal segments. $E$ will be the $4$th dividing point from $F$.)


The resulting figure satisfies all your conditions. $\angle BDC$ is a right angle. The area of $\triangle ABC$ is 18. The area of $\triangle BEC$ is 8.
But notice that the only restriction on $D$ is that it lies somewhere on an arc that extends all the way to $B$. That means we could choose $D$ as close to $B$ as we like.
Consider what happens when $D$ is very close to $B$. The area of $\triangle BDC$ would be very small - much smaller than the area of $8$ required for $\triangle BEC$. In that case $E$ would have to be above $D$. You would still have a quadrilateral $\square BDCE$, but the corners $D$ and $E$ would be opposite what is shown in our diagrams. 
But further, for some choice $D$, $E$ and $D$ will coincide. That is, $\triangle BDC$ will have area $8$, so the point chosen for $E$ will have to be $D$. In this case, the "quadrilateral" $\square BDCE$ will have area $0$.
So we have a range of values for the area of the quadilateral, depending on the choice of $D$.

The same construction can be done regardless of the size of the angle at $B$, but if it is acute, then the arc leaves $\triangle ABC$ before it gets to $B$ (similar to what happens at $C$ above), so it is no longer possible to choose $D$ as close to $B$ as you like. However, it is still true that the area of $\square BDCE$ varies with the choice of $D$. 
Therefore to have an answer to this question, an additional assumption must be made that requires $D$ to be some specific point on the arc. If we assume that $\overline{BC} \perp \overline{AF}$, then there is only one $D$ where this will hold: Where the perpendicular to $\overline{BC}$ passing through $A$ intersects the circle. If $\angle ABC$ is obtuse, then they never intersect, and if it is right, then $D = B = F$ and the whole thing is singular. Thus $D$ can only exist when $\triangle ABC$ is acute. But even in this case, we could consider moving $A$ back and forth on a line parallel to $\overline{BC}$, up to the limits required for acuteness. As $A$ approaches the limit making $\angle B$ right, we once again get $D$ being arbitrarily close to $B$, and thus evidently can have varying values for the area of $\square BDCE$.
So just requiring $\overline{BC} \perp \overline{AF}$ is still not enough for your question to have an answer. I suspect therefore we are meant to assume both $AB = AC$ and $\overline{BC} \perp \overline{AF}$. Then the problem has a solution.
