Prove two vectors are parallel Let $ABCD$ be a parallelogram and $P$, $Q$ two points so that $\overrightarrow{PC} = \frac{1}{3}\overrightarrow{AC}$ and $\overrightarrow{BQ} = 2\overrightarrow{QD}$. 
Find $\alpha$, $\beta$ $\in \mathbb{R}$ with the following property: $\overrightarrow{AP} + \overrightarrow{BQ} = \alpha\overrightarrow{AB} + \beta\overrightarrow{AD}$ and prove that $PQ$ and $BA$ are parallel lines.
For the first part of this problem I found $\alpha = \frac{4}{3}$ and $\beta = \frac{8}{3}$. However, I can't solve the last part. All I know is that I have to find a $\lambda \in \mathbb{R}$ so that $\overrightarrow{PQ} = \lambda\overrightarrow{AB}$, but I can't figure out how to find it.
 A: Representing vectors by $\vec{AB},\vec{AD}$ should help.
Let $\vec{AB}=\vec b,\vec{AD}=\vec d$.
Then, since $\vec{AC}=\vec b+\vec d$, we get
$$\vec{PC}=\frac 13\vec{AC}\implies \vec{AC}-\vec{AP}=\frac 13\vec{AC}\implies \vec{AP}=\frac 23\vec b+\frac 23\vec d$$
and
$$\vec{BQ}=2\vec{QD}\implies \vec{AQ}-\vec{AB}=2(\vec{AD}-\vec{AQ})\implies \vec{AQ}=\frac 13\vec b+\frac 23\vec d$$
Now, it seems that you have some errors in the first part : 
$$\vec{AP}+\vec{BQ}=\frac 23\vec b+\frac 23\vec d+\frac 13\vec b+\frac 23\vec d-\vec b=\alpha\vec b+\beta\vec d\implies \alpha=0,\beta=\frac 43$$
For the second part,
$$\vec{PQ}=\vec{AQ}-\vec{AP}=\frac 13\vec b+\frac 23\vec d-\frac 23\vec b-\frac 23\vec d=-\frac 13\vec b=-\frac 13\vec{AB}$$
A: $\overrightarrow {AC} =  \overrightarrow {AB} + \overrightarrow {AD}\\
\overrightarrow {AP} = \frac 23 \overrightarrow {AC}\\
\overrightarrow {BD} = -\overrightarrow {AB} + \overrightarrow {AD}\\
\overrightarrow {BQ} = \frac 23 \overrightarrow {BD}$
$\overrightarrow {AP}+\overrightarrow {BQ} =$$ \frac 23 \overrightarrow {AC} + \frac 23 \overrightarrow {BD}\\
\frac 23 \overrightarrow {AB} + \frac 23\overrightarrow {AD} -\frac 23\overrightarrow {AB} + \frac 23\overrightarrow {AD}\\
\frac 43 \overrightarrow {AD}$
$\overrightarrow {PQ} = $$\overrightarrow {AB} + \overrightarrow {BQ} - \overrightarrow {AP}\\\overrightarrow {AB}  -\frac 23 \overrightarrow {AB}+\frac 23\overrightarrow {AD}-\frac 23 \overrightarrow {AB} - \frac 23\overrightarrow {AD}\\- \frac 13\overrightarrow {AB}$
