Problem source

The points $P$ and $R$ lie on the sides $AB$ and $AD,$ respectively, of the parallelogram $ABCD.$ The point $Q$ is the fourth vertex of the parallelogram $APQR.$ Prove that $BR,CQ$ and $DP$ meet in
a point.

Solution source

Let $\overrightarrow{AX} = \mathbf{x}$ for all points, so:
\begin{align*}
\mathbf{p} &= p\mathbf{b}\\
\mathbf{r} &= r\mathbf{d}\\
\mathbf{q} &= \mathbf{p}+\mathbf{r} \\
&= p\mathbf{b} + r\mathbf{d}
\end{align*}

Therefore 
\begin{align*}
BR: && \mathbf{b} + \lambda(\mathbf{r}-\mathbf{b}) \\
&&= (1-\lambda) \mathbf{b}+ \lambda r \mathbf{d}  \\
CQ: && \mathbf{c} + \mu(\mathbf{q} - \mathbf{c}) \\
&&= \mathbf{b}+\mathbf{d} + \mu(p\mathbf{b}+r\mathbf{d} - (\mathbf{b}+\mathbf{d}) ) \\
&&= (1+\mu(p-1))\mathbf{b} + (1+\mu(r-1))\mathbf{d} \\
DP: && \mathbf{d} + \nu (\mathbf{p} - \mathbf{d}) \\
&&= \nu p\mathbf{b} +(1-\nu) \mathbf{d}
\end{align*}

So we need $1-\nu = \lambda r, \nu p = 1-\lambda, $ so lets say $1 = \nu + \lambda r, 1 = \lambda + \nu p \Rightarrow \lambda(pr-1) = p-1 \Rightarrow \lambda = \frac{p-1}{pr-1}$ so they intersect at $\frac{rp-r}{pr-1} \mathbf{d} + \frac{pr-p}{pr-1}\mathbf{b}$. If we take $\mu = -\frac{\lambda}{p-1} = 1-pr$ this is clearly also on $CQ$ hence they all meet at a point