4. We have a triangle $\triangle A B C$ and two points $D$ and $E$ on the line segment $B C$, such that $\angle B A D=\angle D A E=\angle E A C$. Furthermore, $|A B|=6,|B D|=4,|D E|=2$ and $|E A|=3$. See the picture below: note that it is not drawn to scale!

What is the length of the line segment $C E$?

![](https://cdn.mathpix.com/cropped/2024_04_17_f4c54c3a9ff27948c305g-2.jpg?height=289&width=508&top_left_y=204&top_left_x=1251)
A) 2
B) $2 \frac{1}{4}$
C) $2 \frac{1}{2}$
D) $2 \frac{3}{4}$
E) 3
A4. E) 3 Name the three equal angles $\alpha$. We see that $|E B|=|E D|+|D B|=2+4=$ $6=|A B|$, and thus triangle $\triangle A B E$ is an isosceles triangle. From this it follows that $\angle A E B=$ $\angle B A E=2 \alpha$ and due to the straight angle at $E$ we find that $\angle A E C=180^{\circ}-2 \alpha$. Since $\angle C A E=\alpha$ and the sum of the angles in $\triangle A C E$ is equal to $180^{\circ}$, it also follows that $\angle A C E=\alpha$. We have thus found another isosceles triangle, namely $\triangle A C E$, and from this it follows that $|C E|=|A E|=3$.