For exercises, 14-18 write paragraph proofs.

Given: parallelogram ABCD, $\overline{\mathrm{AN}}$bisects $\angle \mathrm{DAB}$; $\overline{\mathrm{CM}}$bisects $\angle \mathrm{BCD}$.

Prove: AMCN is a parallelogram.

The given diagram is:

It is being given that ABCD is a parallelogram.

In a parallelogram, both pairs of opposite sides are congruent and parallel.

Therefore, in the parallelogram ABCD, both pairs of opposite sides are congruent and parallel and both pairs of opposite angles are congruent.

Therefore, $\overline{AB}\cong \overline{CD}$, $\overline{AD}\cong \overline{BC}$, $\overline{AB}\parallel \overline{CD}$and $\overline{AD}\parallel \overline{BC}$, $\angle BCD\cong \angle DAB$and $\angle ADC\cong \angle CBA$.

Therefore, $AD=BC$, $\overline{AB}=\overline{CD}$,$\angle BCD=\angle DAB$ and $\angle ADC=\angle CBA$.

It is also being given that $\overline{AN}$bisects $\angle DAB$and $\overline{CM}$bisects $\angle BCD$.

As, $\overline{AN}$bisects $\angle DAB$, therefore by using the definition of angle bisector it can be said that $\angle DAN=\frac{1}{2}\angle DAB$.

As, $\overline{CM}$bisects $\angle BCD$, therefore by using the definition of angle bisector it can be said that $\angle BCM=\frac{1}{2}\angle BCD$.

Therefore, it can be noticed that:

$\begin{array}{c}\angle BCD=\angle DAB\\ \frac{1}{2}\angle BCD=\frac{1}{2}\angle DAB\\ \angle BCM=\angle DAN\end{array}$

Therefore, $\angle BCM\cong \angle DAN$.

In the triangles $△DAN$and $△BCM$, it can be noticed that $\angle BCM\cong \angle DAN$, $AD=BC$and $\angle ADC=\angle CBA$.

Therefore, the triangles $△DAN$and $△BCM$are congruent by ASA congruence.

Therefore, by corresponding parts of congruent triangles, it can be said that $AN\cong CM$and $DN\cong BM$.

As, $\overline{AB}=\overline{CD}$, therefore it can be obtained that:

$\begin{array}{c}\overline{AB}=\overline{CD}\\ \overline{AB}-\overline{BM}=\overline{CD}-\overline{BM}\\ \overline{AB}-\overline{BM}=\overline{CD}-\overline{DN}\left(\because \overline{BM}=\overline{DN}\right)\\ \overline{AM}=\overline{NC}\end{array}$

Therefore, it can be noticed that $AN\cong CM$and $AM\cong NC$.

If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

As, $AN\cong CM$and $AM\cong NC$, therefore, the quadrilateral AMCN is a parallelogram.