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.


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

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

  • $\begingroup$ Oh! Okay. I will try again. But it seems to me that there is something lackage in the question, inadequate information maybe. $\endgroup$ Feb 6 '19 at 12:05
  • 1
    $\begingroup$ @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. $\endgroup$ Feb 6 '19 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. $\endgroup$ Feb 6 '19 at 12:35
  • $\begingroup$ @AnirbanNiloy Updated with solution $\endgroup$ Feb 6 '19 at 13:01
  • 1
    $\begingroup$ @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 $\endgroup$ Feb 6 '19 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.