Let ABCD be a parallelogram and let M be a point on side AB. Let P be the intersection of side BC with the parallel line to AC that passes through M. Let Q be the intersection of AC with the parallel line to DM that passes through B. Show that D, P, and Q are collinear.
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We can prove the theorem in this equivalent form: if $Q$ is the intersection between $AC$ and $PD$, then $BQ$ is parallel to $DM$.
Let $N$ be the intersection between $BQ$ and $CD$, then by similar triangles we have: $$ {CQ\over CA}={CP\over AD}={CN\over AB}. $$ It follows that: $$ CN=AB\cdot{CP\over AD}=AB\cdot{CP\over CB}=AM, $$ where the last equality follows from the intercept theorem applied to parallel lines $PM$ and $AC$.
We get then that triangles $CBN$ and $ADM$ are equal. In particular: $\angle CBN=\angle ADM$ so that $BN\parallel DM$.
Affine maps preserve collinearity, hence we may assume that the original parallelogram $ABCD$ is a square, then embed the construction in $\mathbb{R}^2$ by assuming $A=(0,0)$, $B=(1,0)$, $C=(1,1)$ and $D=(0,1)$. Given $M=(\lambda,0)$, we have $P=(1,1-\lambda)$ and $$ Q=\left(\frac{1}{1+\lambda},\frac{1}{1+\lambda}\right).$$ The collinearity of $D,P,Q$ is a straightforward consequence of $$ P+\lambda D = (\lambda+1) Q,$$ giving that $Q$ is a convex combination of $P$ and $D$. A simple alternative is just to apply Ceva's theorem to the $CBD$ triangle, proving that $AC,BQ$ and $DP$ concur.
By introducing few extra points, you may also prove the same by symmetry: