Chapter 4.7, Problem 11CE

To determine

To prove: Each pair are congruent of the given triangles.

Explanation of Solution

Given information:

  

Plane $M$ is the perpendicular bisecting plane of $\overline{AB}$ at $O$ that is, $M$ is the plane that is perpendicular to $\overline{AB}$ at its midpoint, O.

Points $C$ and $D$ also lie in plane $M$

The three pairs of congruent triangles

Calculation:

Since $O$ is the midpoint of $\overline{AB}$

Therefore, $\overline{AO}=\overline{OB}$

Points $C$ and $D$ lie in plane $M$ which is perpendicular to $\overline{AB}$ at its midpoint, 0.

Therefore,

  $\begin{array}{l}\angle AOC=\angle BOC\\ =\angle AOD=\angle BOD\\ ={90}^{\circ }\end{array}$

  $\begin{array}{l}\text{In }△AOC\text{ and }△BOC\\ \overline{AO}=\overline{OB}\\ \angle AOC=\angle BOC\\ {90}^{\circ }\\ \overline{OC}=\overline{OC}\end{array}$

Therefore, $\text{Δ}AOC\cong \text{Δ}BOC$ by Side-Angle-Side postulate.

  $\begin{array}{l}\text{In Δ}AOD\text{ and Δ}BOD\\ \overline{AO}=\overline{OB}\\ \angle AOD=\angle BOD\\ ={90}^{\circ }\\ \overline{OD}=\overline{OD}\end{array}$

Therefore, $\text{Δ}AOD\cong \text{Δ}BOD$ by Side-Angle-Side postulate.