Chapter 8.3, Problem 15PSB

To determine

To prove : $\overleftrightarrow{HM}$ is parallel to $\overleftrightarrow{JK}$ .

Explanation of Solution

Given information : Here, we have given that $\frac{GJ}{HK}=\frac{GK}{GM}$ and $\angle 1\cong \angle G$

Proof:

As, $\angle 1\cong \angle G$

  $\Rightarrow \overline{GH}\cong \overline{HK}$ (sides opposite to equal angles are equal)

So, $\frac{GJ}{HK}=\frac{GK}{GM}$ will become $\frac{GJ}{GH}=\frac{GK}{GM}$

Now, In $\Delta GJK$ and $\Delta GHM$

  $\frac{GJ}{GH}=\frac{GK}{GM}$ (Proved above)

  $\angle G\cong \angle G$ (Common)

  $\Rightarrow \Delta GJK\sim \Delta GHM$ (By SAS Similarity Rule)

  $\therefore \angle GHM\cong \angle GJK$ (Corresponding angles of similar triangles are equal)

  $\Rightarrow \overleftrightarrow{HM}\parallel \overleftrightarrow{JK}$ (Corresponding angles are equal when lines are parallel)