Final Round

As an intermediate step, we first show that triangles OCM and OCN are congruent. Since $AD$ and $BC$ are parallel, we have (F angles): $\angle CMN=\angle DAM=\frac{1}{2}\angle DAB$. Since $DN$ and $AB$ are parallel, we have (Z angles): $\angle CNM=\angle NAB=\frac{1}{2}\angle DAB$. It follows that $\angle CMN=\angle CNM$, so triangle $CMN$ is isosceles with apex $C$. We obtain $|CM|=|CN|$. Line segments $OC$, $ON$, and $OM$ are radii of the same circle, and therefore of equal length. Triangles $OCM$ and $OCN$ are therefore congruent (three pairs of equal sides).

To show that $\angle OBC=\angle ODC$, we will show that triangles $OBC$ and $ODN$ are congruent. We will do this using the ZHZ-criterion. We will show that $\angle OND=\angle OCB$, and $|ON|=|OC|$, and $|DN|=|BC|$.

The equality $|ON|=|OC|$ follows since $ON$ and $OC$ are radii of the same circle. In part (a), we saw triangles $OCM$ and $OCN$ are congruent. Furthermore, these two triangles are isosceles ($|OC|=|OM|$ and $|OC|=|ON|$). Hence, the four base angles $\angle ONC$, $\angle OCN$, $\angle OMC$, and $\angle OCM$ are equal. We see that $\angle OND=\angle OCB$. The only thing we still need to show is that $|DN|=|BC|$.

Observe that $\angle BMA=\angle DAM$ (Z angles) and $\angle DAM=\angle MAB$ (as $AM$ is the angular bisector of $A$). We find that $\angle BMA=\angle MAB$. Triangle $AMB$ is therefore isosceles and we have $|AB|=|BM|$. We previously saw that $|CM|=|CN|$, and we also have $|AB|=|CD|$ as $ABCD$ is a parallelogram. We therefore obtain $$|DN|=|CD|+|CN|=|AB|+|CM|=|BM|+|CM|=|BC|,$$ which concludes the proof.