ABCD and A_0B_0C_0D_0 are parallelograms, and the points…

$ABCD$ and $A_0B_0C_0D_0$ are parallelograms, and the points $M,N,P,Q$ lie on the segments $AA_0,BB_0,CC_0,DD_0$ so that $\frac{AM}{AA_0}=\frac{BN}{BB_0}=\frac{CP}{CC_0}=\frac{DQ}{DD_0}=\lambda$. $1^\circ$ Show that $MNPQ$ is a parallelogram. $2^\circ$ What is the locus of the intersection of its diagonals as $\lambda$ varies?

Step 1: Position vectors

\[ \vec m=(1-\lambda)\vec a+\lambda\vec a_0,\quad \vec n=(1-\lambda)\vec b+\lambda\vec b_0,\quad \vec p=(1-\lambda)\vec c+\lambda\vec c_0,\quad \vec q=(1-\lambda)\vec d+\lambda\vec d_0. \]

Step 2: $1^\circ$ Opposite sides

\[ \overrightarrow{MN}=(1-\lambda)\overrightarrow{AB}+\lambda\overrightarrow{A_0B_0},\qquad \overrightarrow{QP}=(1-\lambda)\overrightarrow{DC}+\lambda\overrightarrow{D_0C_0}. \]

Since $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{A_0B_0}=\overrightarrow{D_0C_0}$: $\overrightarrow{MN}=\overrightarrow{QP}$.

In the same way

\[ \overrightarrow{MQ}=(1-\lambda)\overrightarrow{AD}+\lambda\overrightarrow{A_0D_0}=(1-\lambda)\overrightarrow{BC}+\lambda\overrightarrow{B_0C_0}=\overrightarrow{NP}. \]

So $MNPQ$ is a parallelogram. (It can be degenerate, i.e. flattened into a segment, if $\overrightarrow{MN}\parallel\overrightarrow{MQ}$.)

Step 3: $2^\circ$ Centre of $MNPQ$

The diagonals bisect each other, so their intersection $S$ is the midpoint of $MP$: \[\vec s=\frac{\vec m+\vec p}{2}=(1-\lambda)\frac{\vec a+\vec c}{2}+\lambda\frac{\vec a_0+\vec c_0}{2}=(1-\lambda)\vec g+\lambda\vec g_0,\] where $G$ and $G_0$ are the centres of the two given parallelograms. Thus $\overrightarrow{GS}=\lambda\,\overrightarrow{GG_0}$.

Because $M$ lies on the segment $AA_0$, $0\le\lambda\le1$, so $S$ runs from $G$ ($\lambda=0$) to $G_0$ ($\lambda=1$).

\[\text{Locus of }S:\ \text{the segment }GG_0\ \ (\text{a single point if }G=G_0)\]