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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?

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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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  • $\begingroup$ Oh! Okay. I will try again. But it seems to me that there is something lackage in the question, inadequate information maybe. $\endgroup$ – Anirban Niloy Feb 6 at 12:05
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    $\begingroup$ @AnirbanNiloy No, all the information is there. Normally, this construction would generate a trapezoid $ACEF$, only in very special circumstances would it be parallelogram. You can use them to find out the height of $ACEF$. If you still can't figure it out, I can write out the full solution. $\endgroup$ – Todor Markov Feb 6 at 12:20
  • $\begingroup$ I tried so much. Nevertheless, I couldn't figure it out. Besides, I proved the $ACEF$ as a parallelogram by following the above condition. But when it would be a trapezoid? To my novice mind, it seems to me really confused. $\endgroup$ – Anirban Niloy Feb 6 at 12:35
  • $\begingroup$ @AnirbanNiloy Updated with solution $\endgroup$ – Todor Markov Feb 6 at 13:01
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    $\begingroup$ @AnirbanNiloy If you take a random triangle $ABC$, then rotate it and align it to get $CDE$, $AC$ will not be parallel to $DE$ unless $ACB$ is a right angle. This follows from angles formed by parallel lines, see mathnstuff.com/math/spoken/here/2class/260/trans.htm $\endgroup$ – Todor Markov Feb 6 at 13:21

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