Chapter 3.3, Problem 22PSC

To determine

To Prove: $\angle MHO$ is larger than $\angle K$

Explanation of Solution

Given information : In this question, we have given that $\overline{GJ}$ and $\overline{KH}$ bisect each other.

Formula used: ASA Congruency rule.

Proof: As, $\overline{GJ}$ and $\overline{KH}$ bisect each other,$\Rightarrow \overline{KM}=\overline{MH}$ and $\overline{JM}=\overline{GH}$

Now, In $\Delta KMG$ and $\Delta JHM$ ,

  $\overline{KM}=\overline{MH}$ (Proved above)

  $\angle KMG\cong \angle JMH$ (Vertically Opposite Angles)

  $\overline{JM}=\overline{GH}$ (Proved above)

  $\therefore \Delta KMG\cong \Delta JHM$ (BY ASA Congruency rule)

  $\therefore \angle GKM\cong \angle JHM$ (BY CPCT)

Now, as we are adding $\angle JHO$ in $\angle JHM$ to make it $\angle MHO$ .

  $\angle GKM\cong \angle JHM$

If we add something on either side of any equation, the added side will always become greater.

  $\Rightarrow \angle MHO>\angle K$

  $\therefore$ $\angle MHO$ is larger than $\angle K$