Chapter 5.5, Problem 9PSA

To determine

To prove:

  $\overrightarrow{RO}$ bisects angle MRH.

  $\overrightarrow{OR}$ bisects angle MOH.

  $\overrightarrow{MH}$ bisects angle OMR.

  $\overrightarrow{HM}$ bisects angle RHO.

Explanation of Solution

Given information:

MOHRis a rhombus.

Proof:

  

It is given that, MOHRis a rhombus.

All sides of a rhombus are congruent.

  $\overline{MO}\cong \overline{OH}\cong \overline{HR}\cong \overline{MR}$

If sides are congruent then corresponding angles are also congruent.

  $\begin{array}{l}\angle 3\cong \angle 8\\ \angle 4\cong \angle 7\end{array}$

Opposite sides of rhombus are parallel.

  $\begin{array}{l}\overline{RM }\parallel \overline{HO }\\ \overline{MO }\parallel \overline{RH }\end{array}$

If lines are parallel then their alternate interior angles are congruent.

  $\begin{array}{l}\angle 3\cong \angle 7\\ \angle 4\cong \angle 8\end{array}$

By transitive property, we get

  $\begin{array}{l}\angle 3\cong \angle 4\\ \angle 7\cong \angle 8\end{array}$

By definition of angle bisector, we get

  $\overrightarrow{RO}$ bisects angle MRH.

  $\overrightarrow{OR}$ bisects angle MOH.

If sides are congruent then corresponding angles are also congruent.

  $\begin{array}{l}\angle 2\cong \angle 5\\ \angle 1\cong \angle 6\end{array}$

If lines are parallel then their alternate interior angles are congruent.

  $\begin{array}{l}\angle 2\cong \angle 6\\ \angle 1\cong \angle 5\end{array}$

By transitive property, we get

  $\begin{array}{l}\angle 1\cong \angle 2\\ \angle 5\cong \angle 6\end{array}$

By definition of angle bisector, we get

  $\overrightarrow{MH}$ bisects angle OMR.

  $\overrightarrow{HM}$ bisects angle RHO.