Chapter 6.2, Problem 29PPS

To determine

To prove:

Theindicated proof for the following:

  $\overline{\mathrm{E}\mathrm{C}}$ bisects $\overline{\mathrm{A}\mathrm{D}}$

Explanation of Solution

Given information:

  

  $▱\mathrm{A}\mathrm{C}\mathrm{D}\mathrm{E}$ is parallelogram

Proof:

Know that:

  $▱\mathrm{A}\mathrm{C}\mathrm{D}\mathrm{E}$ is parallelogram, that means it’s opposite sides are congruent and parallel as well.

The alternate angles are always congruent when two parallel lines get cut by transversal.

Here, the given that $▱\mathrm{A}\mathrm{C}\mathrm{D}\mathrm{E}$ is parallelogram, so opposite sides $\overline{\mathrm{E}\mathrm{A}}$ and $\overline{\mathrm{E}\mathrm{C}}$ are congruent and parallel as $\overline{\mathrm{E}\mathrm{A}}\cong \overline{\mathrm{D}\mathrm{C}}$ and $\overline{\mathrm{E}\mathrm{A}}\parallel \overline{\mathrm{D}\mathrm{C}}$

Therefore by the definition of parallelogram:

  $\begin{array}{l}\angle \mathrm{A}\mathrm{E}\mathrm{B}\cong \angle \mathrm{D}\mathrm{C}\mathrm{B}\\ \angle \mathrm{E}\mathrm{A}\mathrm{B}\cong \angle \mathrm{C}\mathrm{D}\mathrm{B}\end{array}$  (Because alternate interior angles are always congruent )

Therefore, by using the angle side angel theorem:

  $\mathrm{\Delta }\mathrm{E}\mathrm{B}\mathrm{A}\cong \mathrm{\Delta }\mathrm{C}\mathrm{B}\mathrm{D}$

Know that - corresponding parts of congruent triangle are congruent as:

  $\begin{array}{l}\overline{\mathrm{E}\mathrm{B}}\cong \overline{\mathrm{B}\mathrm{C}}\\ \overline{\mathrm{A}\mathrm{B}}\cong \overline{\mathrm{B}\mathrm{D}}\end{array}$

Therefore, by the definition of segment bisector, get that:

  $\overline{\mathrm{E}\mathrm{C}}$ bisects $\overline{\mathrm{A}\mathrm{D}}$ or $\overline{\mathrm{A}\mathrm{D}}$ bisects $\overline{\mathrm{E}\mathrm{C}}$

Hence proved.