Problem
If the lengths of the two sides $AB$ and $CD$ of a quadrilateral $ABCD$ are equal, then the extensions of these sides and the line connecting the midpoints $M$ and $N$ of the other two sides $AD$ and $BC$ will make equal angles.
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Solution
Let $P$ and $Q$ be the points where the extensions of $BA$ and $CD$ intersect with the line $MN$, respectively.
Furthermore, let $E$ be the midpoint of the diagonal $DB$.
Then, $△ABD$ and $△MED$ share $∠BDA \ (=∠EDM)$,
$$BD∶ED=DA∶DM=2∶1,$$
$$∴ \ △ABD∼△MED,$$
$$∴ \ AB∥ME,$$
$$∴ PB∥ME,$$
$$∴ \ ∠BPN=∠EMN. \qquad [1]$$
$△BCD$ and $△BNE$ share $∠DBC \ (=∠EBN)$,
$$BC∶BN=DB∶EB=2∶1,$$
$$∴ \ △BCD∼△BNE,$$
$$∴ \ DC∥EN,$$
$$∴ \ QC∥EN,$$
$$∴ \ ∠CQN=∠ENM. \qquad [2]$$
However,
$$ME=\frac{1}{2} AB, \qquad NE={1}{2} CD \qquad and \qquad AB=CD,$$
$$∴ \ ME=NE.$$
Therefore, $△EMN$ is an isosceles triangle,
$$∠EMN=∠ENM. \qquad [3]$$
Therefore, from $[1], \ [2]$ and $[3]$,
$$∠BPN=∠CQN.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.43.