Chapter 5.3, Problem 14PSB

To determine

To Prove: $\overline{FH}\cancel{\cong }\overline{MO}$ .

Explanation of Solution

Given:

  

  $\begin{array}{l}\overline{FJ}\cancel{\parallel }\overline{KO}\\ \overline{FH}\parallel \overline{MO}\\ \overline{HK}\cong \overline{MJ}\end{array}$

Concept Used:

When parallel lines is intersected by a transversal corresponding angles are equal.

Proof:

Here, we have

  $\begin{array}{l}\overline{FJ}\cancel{\parallel }\overline{KO}\\ \overline{FH}\parallel \overline{MO}\\ \overline{HK}\cong \overline{MJ}\end{array}$

  $\begin{array}{l}\text{In }△\text{FHJ and }△\text{OKM,}\\ \overline{FJ}\cancel{\parallel }\overline{KO}\\ \Rightarrow \angle J\cancel{\cong }\angle O\\ \overline{FH}\parallel \overline{MO}\\ \Rightarrow \angle H\cong \angle M\\ \because \overline{HK}\cong \overline{MJ}\\ \Rightarrow \overline{HJ}\cong \overline{MK}\\ \text{But }△\text{FHJ }\cancel{\cong }△\text{OKM}\\ \Rightarrow \overline{FH}\cancel{\cong }\overline{MO}\end{array}$