Chapter 5.6, Problem 25PPS

To determine

To prove:The angle $\angle XZU$ is greater than $\angle UZV$ .

Explanation of Solution

Given information:

The given figure is

  

  $\begin{array}{l}\overline{XU}\cong \overline{VM},VW>XW\\ \overline{XU}\Vert \overline{VM}\end{array}$

Formula Used:

Hinge theorem is

If two sides of a triangle correspond to two sides of another triangle, and the angle of inclination of the first triangle is greater than the angle of inclination of the second triangle, so the sides of the first triangle are longer than the sides of the third side of the second triangle.

Prove:

It is given that $\overline{XU}\cong \overline{VM},VW>XWand\overline{XU}\Vert \overline{VM}$ .

From the figure and the given datas.

  $\angle XUZ\cong \angle VWZ$

Similarly, $\angle XUZ\cong \angle VWZ$

Thus, by AAS congruency.

  $\Delta UZX\cong \Delta WZV$

So, $\overline{XZ}\cong \overline{VZ}$ and $\overline{WZ}\cong \overline{WZ}$ .

Now, by Hinge theorem.

  $m\angle WZV>m\angle WZX$  ...... (1)

Similarly $m\angle WZX\cong m\angle UZV$  ...... (2)

Thus, from (1) and (2)

  $m\angle XZU>m\angle UZV$

Hence, proved.