4. In concave quadrilateral $A B C D$, $\angle A=\angle C=$ $40^{\circ}, \angle B=50^{\circ}, E, F, G, H$ are the midpoints of sides $A B, B C$, $C D, D A$ respectively. Then the shape of quadrilateral $E F G H$ is $\qquad$
4. Rectangle.

As shown in Figure 5, connect $AC$ and $BD$. By the Midline Theorem of a triangle, we have
$$
\begin{array}{l}
E H \perp \frac{1}{2} B D, \\
F G \perp \frac{1}{2} B D .
\end{array}
$$

Thus, $E H \perp F G$.
Therefore, quadrilateral $E F G H$ is a parallelogram.
$$
\begin{array}{l}
\text { By } \angle B A D+\angle A B C=90^{\circ}, \\
\angle B C D+\angle A B C=90^{\circ},
\end{array}
$$

we get $A D \perp B C, C D \perp A B$.
Thus, $D$ is the orthocenter of $\triangle A B C$, and $B D \perp A C$. Since $G H \parallel A C$, then $B D \perp G H$.
Therefore, $F G \perp G H$.
Hence, quadrilateral $E F G H$ is a rectangle.