[ Varignon's Parallelogram ]

The diagonals of a convex quadrilateral are equal to $c$ and $d$ and intersect at an angle of $45^{\circ}$. Find the segments connecting the midpoints of opposite sides of the quadrilateral.

#
Let the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ be equal to $c$ and $d$ respectively, intersecting at point $O$ with $\angle AOB = 45^\circ$. If $K, P, M$, and $N$ are the midpoints of the sides $AB, CD, BC$, and $AD$ respectively, then $KM$ and $KN$ are the midlines of triangles $ABC$ and $BAD$, so

\[
\begin{gathered}
KM \| AC, \quad KM = \frac{1}{2} AC = \frac{1}{2} c \\
KN \| BD, \quad KN = \frac{1}{2} BD = \frac{1}{2} d \\
\angle MKN = \angle AOB = 45^\circ, \quad \angle KMP = 180^\circ - \angle MKN = 135^\circ.
\end{gathered}
\]

From triangles $KMN$ and $KPM$ using the cosine rule, we find that

\[
\begin{gathered}
MN^2 = KM^2 + KN^2 - 2 \cdot KM \cdot KN \cdot \cos 45^\circ = \frac{1}{4} c^2 + \frac{1}{4} d^2 - \frac{1}{4} cd \sqrt{2} = \frac{1}{4} (c^2 + d^2 - cd \sqrt{2}) \\
KP^2 = MK^2 + MP^2 - 2 \cdot MK \cdot MP \cdot \cos 135^\circ = \frac{1}{4} c^2 + \frac{1}{4} d^2 + \frac{1}{4} cd \sqrt{2} = \frac{1}{4} (c^2 + d^2 + cd \sqrt{2})
\end{gathered}
\]

Therefore,

\[
MN = \frac{1}{2} \sqrt{c^2 + d^2 - cd \sqrt{2}}, \quad KP = \frac{1}{2} \sqrt{c^2 + d^2 + cd \sqrt{2}}
\]

## Answer

\[
\frac{1}{2} \sqrt{c^2 + d^2 \pm cd \sqrt{2}}
\]