Problem
Of the two points of intersection between the two lines that trisect $∠B$ of an isosceles triangle $ABC$ and the perpendicular line $AD$ drawn from the vertex $A$ to the base $BC$, the one closer to $A$ is called $M$ and the other is called $N$.
If the intersection point of the line $CN$ and the side $AB$ is called $E$, then the line $EM$ is parallel to the line $BN$.
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Solution
Since $AD$ is the axis of symmetry for $△ABC$, if $BM$ and $BN$ trisect $∠B$, then $CM$ and $CN$ also trisect $∠C$, and
$$∠EBM=\frac{1}{3}∠B, \qquad ∠ECM=\frac{1}{3}∠C \qquad and \qquad ∠B=∠C,$$
$$∴ \ ∠EBM=∠ECM.$$
Therefore, $B,\ C, \ M$ and $E$ are on the same circumference.
Thus,
$$∠BME=∠BCE=\frac{1}{3}∠C \qquad and \qquad ∠MBN=\frac{1}{3}∠B,$$
$$∴ \ ∠BME=∠MBN,$$
$$∴ \ EM∥BN.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.35.