Area of triangle interior to parallelogram

Question

Hi you can help me with this exercise? I have to find the area of the triangle $QOP$ in terms of the parallelogram $ABCDB$ but I do not know how to prove that the area of the triangle $COD$ is $1/8$ of $ABCD$ area. Thanks for you help.

In parallelogram $ABCD$ of the diagram the line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P$. From vertex $C$ line $CQ$ is drawn bisecting side $AD$ at $M$ at meeting $AB$ (extended) at $Q$. Lines $DP$ and $CQ$ meet and point $O$. If the area of parallelogram is $k$ show that area of triangle $COD$ is $\frac{1}{8}k$.

Answer 1

Connect $M$ and $N$ by a line. By construction, the height of the parallelogram $MNCD$ is $\frac{h}{2}$, where $h$ is the height of $ABCD$. Now consider that $O$ is the center of $MNCD$, since it is the crossing point of its two diagonals. Therefore, the height of triangle $MON$ and that of triangle $COD$ are both equal to $\displaystyle \frac{h}{4}$. Calling $b$ the length of the segments $AB=MN=DC$, the area of $COD$ is then given by $\displaystyle \frac{b \cdot \frac{h}{4}}{2}=\frac{bh}{8}=\frac{k}{8}$.

Answer 2

Triangle $QAM$ is congruent to $CDM$, and $BNP$ is congruent to $CND$, by ASA. Draw segment $MN$, so triangle $MON$ is congruent to $COD$. Triangle $QOP$ is similar to $MON$, with an edge ratio of 3:1, so it has 9 times the area. Notice that the area of triangle $QOP$ is the area of parallelogram $ABCD$ plus the area of $COD$ (imagine rotating $QAM$ up about point $M$, and $BNP$ up about point $N$, they will overlap $ABCD$, and cover $COD$ twice).

So we have $QOP = ABCD + COD = k + x = 9x$, so $x = \frac{1}{8}k$.