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Q15

Expert-verifiedFound in: Page 175

Book edition
Student Edition

Author(s)
Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages
227 pages

ISBN
9780395977279

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.

It is proved that the quadrilateral *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 $\u25b3DAN$ and $\u25b3BCM$, it can be noticed that $\angle BCM\cong \angle DAN$, $AD=BC$ and $\angle ADC=\angle CBA$.

Therefore, the triangles $\u25b3DAN$ and $\u25b3BCM$ 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:

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

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.

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