Chapter 5.5, Problem 2PSA

Solution

1.

To find: The reason of being EFHJ is said to be a parallelogram

Given:

EFHJ is a parallelogram

  $\angle 1\cong \angle 2$

Explanation:

It is given in the question that EFHJ is a parallelogram.

2.

To find: The reason of being $\angle J$ is congruent to $\angle F$

Given:

EFHJ is a parallelogram

  $\angle 1\cong \angle 2$

Explanation:

  $\angle J$ is congruent to $\angle F$ because opposite angles of parallelogram are equal and congruent.

3.

To find: The reason of being $\overline{JH}$ is congruent to $\overline{EF}$

Given:

EFHJ is a parallelogram

  $\angle 1\cong \angle 2$

Explanation:

  $\overline{JH}$ is congruent to $\overline{EF}$ because opposite sides of parallelogram are equal and congruent.

4.

To find: The reason of being $\angle 1$ is congruent to $\angle 2$

Given:

EFHJ is a parallelogram

  $\angle 1\cong \angle 2$

Explanation:

It is given in the question that $\angle 1$ is congruent to $\angle 2$

5.

To find: The reason of being $\Delta KJH$ is congruent to $\Delta GFE$

Given:

EFHJ is a parallelogram

  $\angle 1\cong \angle 2$

Concept used:

Here, we are using the concept of Angle Side Angle property.

Explanation:

Here, In $\Delta KJH$ and $\Delta GFE$

  $\angle J=\angle F$ (Proved above)

  $JH=EF$ (Opposite sides of parallelogram are equal)$\angle 1=\angle 2$ (given)

Hence, $\Delta KJH\cong \Delta GFE$

(by using Angle Side Angle property)

6.

To find: The reason of being $\overline{KH}$ is congruent to $\overline{EG}$

Given:

EFHJ is a parallelogram

  $\angle 1\cong \angle 2$

Explanation:

Here, In $\Delta KJH$ and $\Delta GFE$

  $\angle J=\angle F$ (Proved above)

  $JH=EF$ (Opposite sides of parallelogram are equal)$\angle 1=\angle 2$ (given)

Hence, $\Delta KJH\cong \Delta GFE$

(by using Angle Side Angle property)

So, $\overline{KH}\cong \overline{EG}$

(by corresponding parts of congruent triangles)