Chapter 5, Problem 19CT

To determine

To prove: $\angle 1$ and $\angle 2$ are complementary.

Explanation of Solution

Given:

The diagram:

$\overline{PR}\parallel \overline{VO}$ ; $\overline{RO}\parallel \overline{PV}$ ; $\overline{PR}\cong \overline{RO}$

Concept Used:

Theorem 3-2: Alternate angles are equal to each other.

Theorem 3-11: The addition of all angles of a triangle is 180.

Theorem 5-13: Rhombus’s diagonals are perpendicular.

Theorem 5-14: Two angles of the rhombus bisected by the each diagonal of a rhombus.

Theorem 5-17: the parallelogram is a rhombus if two consecutive sides of a parallelogram are equal.

Consider the figure shown below:

Consider the quadrilateral $PROV$ .

It is known that if opposite sides of a quadrilateral are parallel, then it is a parallelogram.

Here it is given that, $\overline{PR}\parallel \overline{VO}$ and $\overline{RO}\parallel \overline{PV}$ so, $PROV$ is a parallelogram.

Also, it is given that $\overline{PR}\cong \overline{RO}$ , so by theorem 5-17, the parallelogram $PROV$ is a rhombus.

By theorem 5-13, $\overline{RV}\perp \overline{PO}$

So, $\angle REO=90°$

Consider the triangle $△REO$ ,

So, by theorem 3-11,

  $\begin{array}{c}m\angle REO+m\angle EOR+m\angle ORE=180\\ 90+m\angle EOR+m\angle 2=180\\ m\angle EOR+m\angle 2=180-90\end{array}$

$m\angle EOR+m\angle 2=90$ …… (1)

Also, it is given that $\overline{PR}\parallel \overline{VO}$ and $\overline{RV}$ is a transversal, so by theorem 3-2,

  $\begin{array}{l}\angle PRV\cong \angle RVO\\ \angle PRV\cong \angle 1\end{array}$

That is,

$m\angle PRV=m\angle 1$ ….. (2)

Also, by theorem 5-14, the diagonal $\overline{RV}$ bisects the angle $\angle PRO$ .

That is,

  $m\angle PRV=m\angle VRO$ ….. (3)

By using (1), (2) and (3), it can be said that,

  $m\angle 1+m\angle 2=90$

Thus, $\angle 1$ and $\angle 2$ are complementary.