<s>[INST] In $\triangle ABC$, $D$ is an exterior point such that $AC = CD$ and $CE$ is parallel to $AF$. Find the area of $ABDF$. 
In $\triangle ABC$, $CB$ is extended upto $D$ so that $AC$ = $CD$. An angle $\angle DCE$ is drawn at point $C$ so that is equal to $\angle CAB$ and $AB$ meets $CE$ at $I$.$E$ is such an external point that satisfies the term $\angle ACB$ = $\angle CDE$. The line parallel to $CE$ is drawn from $A$ meets extended $DE$ $F$. $AFEC$ is a parallelogram. $AB$ = $4IB$ and the area of $\triangle ABC$ is $\frac{9}{4}$$\sqrt{ 15}$. What is the area of $ABDF$?


What I try:
Here, $\triangle ABC$ $\cong$ $\triangle CDE$ (with the above condition). So the area of $\triangle CDE$ will be equal to $\frac{9}{4}$$\sqrt {15}$
Let denote the height of $\triangle ABC$ = $h$
As, $AI:IB$ = 4:1 and $h$ of both $\triangle ACI$ and $CIB$ are equal.
So, the area of $\triangle ACI$ = $\frac{4}{5}$×$\frac{9}{4}$$\sqrt {15} $= $\frac{9}{5}$$\sqrt { 15}$
Thus, the area of $BIED$ will also be equal to $\frac{9}{5}$$\sqrt {15}$, because $\triangle CIB$ is the common segment of both the triangle $\triangle ACB$ and $\triangle CDE$.
But, I got stuck whenever I try to find out the area of parallelogram of $AFEC$. I can't figure out even the length and height of paralleologram. I think that some parts of the diagram need to be showed as congruent. So, I can find the area of $ACEF$ somehow.
But I failed. How can I do this and how many ways it can be solved?
 [/INST] Hint: You forgot to use the fact that $AFEC$ is a parallelogram. It doesn't directly follow from the rest of the statement, but is rather very informative.
Solution:
We have $\angle ACB = \angle CDE$. Since $AFEC$ is a parallelogram, $AC$ is parallel to $FD$, therefore, $\angle ACD + \angle CDE = 180^\text{o}$. From these two we can conclude that $\angle ACB$ is a right angle. Thus $CD$ is a height of $AFEC$
$S_{ABDF} = S_{ACFE} = AC \times CD = AC^2$ (both are $S_{ACDE} - S_{ABC}$).
So we just need to find $AC$. For the right triangle $ABC$, we have $\frac{AC}{AB} = \frac{AI}{AC}$ (proof), so $AC^2 = AB \times AI$. Similarly $BC^2 = AB \times BI$, so $\frac{AC^2}{BC^2} = \frac{AI}{BI} = 3$.
Recall $AC \times BC = \frac{9}{2}\sqrt{15}$. Since $AC = \sqrt{3}BC$, $AC \times BC = \sqrt{3}BC^2 = \frac{9}{2} \times \sqrt{15}$, so $BC^2 = \frac{9}{2} \sqrt{5}$. and so
$$S_{ABDF} = AC^2 = 3 BC^2 = \frac{27}{2}\sqrt{5}$$
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