Selection Examinations for the IMO, 2013

We prove that the lines $DM$ and $DN$ are the interior and the exterior bisector of the angle $\angle ADC$. Let $X$ be the point of intersection of the lines $AF$ and $CD$.

Menelaus' theorem for the triangle $EBC$ and the collinear points $A, M$ and $F$ says
$$\frac{|EA| \ |BF| \ |CM|}{|AB| \ |FC| \ |ME|} = 1. \qquad (2)$$
Because $AB \parallel CD$, the triangles $AEM$ and $XCM$ have identical angles, hence they are similar and $\frac{|CM|}{|ME|} = \frac{|MX|}{|AM|}$. From $|AE| = |FC|$ and (2) we now derive
$$\frac{|BF|}{|AB|} = \frac{|AM|}{|MX|}. \qquad (3)$$
From $AB \parallel CD$ we get $\angle BAX = \angle DXA$, and from $AD \parallel BC$ we get $\angle AFB = \angle XAD$. The triangles $ABF$ and $XDA$ are thus similar and $\frac{|BF|}{|AB|} = \frac{|AD|}{|DX|}$. From (3) we derive $\frac{|AM|}{|MX|} = \frac{|AD|}{|DX|}$, which means that $DM$ is the interior bisector of $\angle ADX$.

An analogous proof follows for the exterior bisector. Menelaus' theorem for the triangle $GBC$ and the collinear points $N, A$ and $F$ says
$$\frac{|GA| \ |BF| \ |CN|}{|AB| \ |FC| \ |NG|} = 1. \qquad (4)$$
Because $AB \parallel CD$, the triangles $AGN$ and $XCN$ have identical angles, hence they are similar and $\frac{|CN|}{|NG|} = \frac{|NX|}{|AN|}$. From $|AG| = |FC|$ and (4) we now derive
$$\frac{|BF|}{|AB|} = \frac{|AN|}{|NX|}. \qquad (5)$$
From the ratio $\frac{|BF|}{|AB|} = \frac{|AD|}{|DX|}$ we get $\frac{|AN|}{|NX|} = \frac{|AD|}{|DX|}$. Hence, $DN$ is the exterior bisector of the angle $\angle ADX$.

Since the exterior and interior bisector are perpendicular, we have $\angle NDM = 90^\circ$.