Chapter 3.7, Problem 15PSB

To determine

To Prove: $\Delta FHM$ is isosceles

Explanation of Solution

Given information: $\overline{HK}\cong \overline{JM}$ , $\overline{GJ}\cong \overline{JK}$ , $\overline{OK}\cong \overline{JK}$ , $\overline{GJ}$ and $\overline{OK}$ are $\perp$ to $\overline{HM}$ is provided in the question which will help to find the solution.

Proof: Here, $\overline{GJ}=\overline{JK}=\overline{OK}$ (given)

And, $\angle GJH=\angle OKM={90}^{\circ }$

So that, $\overline{HG}=\overline{OM}$ (sides of opposite angles are equal)

And, $\overline{HK}\cong \overline{JM}$

  $\overline{HJ}+\overline{JK}=\overline{JK}+\overline{KM}$

  $\overline{HJ}=\overline{KM}$

Hence, we can say that

  $\Delta FHM$ is isosceles.