Shaded area as a proportion of the parallelogram? I really have tried here.
I am so stuck and it's driving me crazy because I know there's something simple here.
So far I have fixed some lengths, calling $AP$ 2, $PB$ 1, and $DC$ 3, and also recognized that $\delta PBX$ and $\Delta DCX$ are similar triangles.
Please can someone give me a hint for this question?

 A: Hint, as requested:
The shaded triangles are similar. They divide the height of the parallelogram in a ratio of $3:1$
Full solution:
The area of the parallelogram is $bh$.
The area of the larger triangle is $\frac{1}{2}\times b\times\frac{3h}{4}=\frac{3bh}{8}$
The area of the smaller triangle is $\frac{1}{2}\times\frac{b}{3}\times\frac{h}{4}=\frac{bh}{24}$
The combined area of the shaded region is
$$\frac{3bh}{8}+\frac{bh}{24}=\frac{5}{12}bh$$
A: 
Let base, $CD=x$ & altitude, $MN=y$. From $A$-$A$-$A$ similarity, $\Delta CDX\sim\Delta PBX$
$$\frac{XN}{XM}=\frac{CD}{PB}=\frac{x}{x/3}\implies XN=3XM$$
$$XM+XN=y\implies XM+3XM=y\implies XM=\frac y4 \quad \text{&}\quad XN=\frac{3y}{4}$$
Hence the shaded area of parallelogram $$\frac{1}{2}(CD)(XN)+\frac{1}{2}(PB)(XM)=\frac{1}{2}(x)\left(\frac{3y}{4}\right)+\frac{1}{2}\left(\frac{x}{3}\right)\left(\frac{y}{4}\right)=\frac{5}{12}xy$$
$$=\frac{5}{12}(\text{Area of parallelogram ABCD})$$
A: Let [] denote areas and [ABCD] = $1$. Note that BPX and CDX are similar, which yields $\frac{PX}{XC}=\frac{BX}{XD}=\frac13$. Then, the shaded area is
\begin{align}
[\text{Shaded}] &= [BPX] + [CXD] \\
& = \frac{PX}{PC} [BPC]+ \frac{XD}{BD} [BDC] \\
& =\frac14\cdot \frac16 + \frac34\cdot \frac12= \frac5{12}\\
\end{align}
