B4. Triangle $A B C$ is isosceles with apex $C$. The midpoint of $A B$ is called $M$. On segment $C M$ lies a point $D$ with $\frac{|C D|}{|D M|}=\frac{3}{2}$. Line $B D$ intersects segment $A C$ at point $E$.

Calculate $\frac{|C E|}{|E A|}$. (Note: the figure below is not to scale.)

![](https://cdn.mathpix.com/cropped/2024_04_17_b17c3267eb783a5c7d68g-1.jpg?height=420&width=297&top_left_y=2006&top_left_x=1619)
4. base multiplied by the height. If for triangles $B D M$ and $B C D$ the sides $D M$ and $C D$ are taken as the base, then the triangles have the same height. The areas therefore relate as $2: 3$. If the area of triangle $B D M$ is $2 x$, then the area of triangle $B C D$ is $3 x$.

In the same way, the areas of triangles $D E M$ and $C D E$ relate as $2: 3$. The area of $D E M$ is called $2 y$, then the area of $C D E$ is $3 y$. Triangles $B E M$ and $A E M$ also have equal areas, since $M$ is the midpoint of $A B$. Therefore, the area of triangle $A E M$ is $2 x + 2 y$.

Finally, we note that triangles $A C M$ and $B C M$ have equal areas, namely half of triangle $A B C$. So $(2 x + 2 y) + 2 y + 3 y = 2 x + 3 x$, or $7 y = 3 x$.

![](https://cdn.mathpix.com/cropped/2024_04_17_51fa63b20784abd70d25g-2.jpg?height=734&width=471&top_left_y=147&top_left_x=1429)

The total area of $A B C$ is $10 x$. Triangle $B C E$ has an area of $3 x + 3 y = \frac{30}{7} x$. The area of triangle $A B E$ is then $10 x - \frac{30}{7} x = \frac{40}{7} x$. The ratio $\frac{|C E|}{|E A|}$ is thus $\frac{3}{4}$.