Given a quadrilateral $ABCD$, prove that the quadrilateral formed by its midpoints, $EFGH$, is a parallelogram.
[Math] Prove that the quadrilateral formed by connecting the midpoints of a quadrilateral is a parallelogram.
euclidean-geometrygeometryquadrilateral
Related Solutions
Angles in gray triangles are $\alpha, \beta$ and $90^\circ$. Catheti (green and red) are equal, when $\alpha=\beta$.
So, central rectangle can be square when $\alpha=\beta ~~$ too.
The midpoint of $AB$ is $\frac12(A+B)$.
The midpoint of $BC$ is $\frac12(B+C)$.
The vector from the midpoint of $AB$ to the midpoint of $BC$ is $\frac12(B+C)-\frac12(A+B) = \frac12(C-A)$.
Repeat for $AD$ and $DC$ to see that you get the same vector there.
Best Answer
Since EH is a midline of triangle ABD, $EH \parallel BD$. Likewise, $FG \parallel BD$ looking at triangle BCD. Therefore $EH \parallel BD \parallel FG$. Similarly you can show $EF \parallel HG$.