5. In the diagram, $A B C D$ is a parallelogram, $M$ is the midpoint of $A B$ and $X$ is the point of intersection of $A C$ and $M D$.
What is the ratio of the area of $M B C X$ to the area of $A B C D$ ?
Solution
Solution 1
Say that the area of triangle $X A M$ is 1 square unit. Note that $\angle X M A=\angle X D C$ and $\angle X A M=\angle X C D$ (since they are alternate angles between parallel lines), so triangles $X C D$ and $X A M$ are similar. Since $D C=2 A M$, the area of $X C D$ is $1 \times 2^{2}=4$ square units.
Say the area of triangle $D A X$ is $x$ square units. Then, since the area of $A C D$ is twice the area of $A M D$ (they are triangles with equal heights and one base half that of the other), we have $2 \times(1+x)=4+x$, which gives $x=2$ square units.
The area of $A B C D$ is twice that of $A C D$, which is 12 square units.
The area of $M B C X$ is $12-(4+2+1)=5$ square units.
So the desired ratio is $5: 12$.

Solution 2
Note that $\angle X M A=\angle X D C$ and $\angle X A M=\angle X C D$ (since they are alternate angles between parallel lines), so triangles $X C D$ and $X A M$ are similar. Since $D C=2 A M$, the perpendicular from $D C$ to $X$ is twice the perpendicular from $A M$ to $X$.
Hence the perpendicular height of the triangle $A M X$ is one third of the perpendicular height of the parallelogram. Hence the area of $A M X$ is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{3}=\frac{1}{12}$ of the area of the parallelogram.
The area of triangle $A B C$ is clearly half of the area of the parallelogram, so the area of $M B C X$ is $\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$ of the area of the whole parallelogram.
Hence the desired ratio is $5: 12$.