# 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?

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

$$S_{ABDF} = AC^2 = 3 BC^2 = \frac{27}{2}\sqrt{5}$$

• Oh! Okay. I will try again. But it seems to me that there is something lackage in the question, inadequate information maybe. Feb 6 '19 at 12:05
• @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. Feb 6 '19 at 12:20
• 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. Feb 6 '19 at 12:35
• @AnirbanNiloy Updated with solution Feb 6 '19 at 13:01
• @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 Feb 6 '19 at 13:21