Chapter 10.3, Problem 24PPS

To determine

Prove :

$\overline{JP}$ bisects $\overline{KM}$ and arc KM.

Explanation of Solution

Given :

In $\odot P$ , $\overline{KM}\perp \overline{JP}$

Calculation:

Given: In $\odot P$ , $\overline{KM}\perp \overline{JP}$

Consider $\Delta KLP\text{ and }\Delta \text{MLP}$

$\angle KLP\text{ and }\angle \text{MLP}$ are right angles.

So, $\angle KLP\cong \angle MLP\mathrm{............}[\text{definition of congruency]}$

$\overline{LP}\cong \overline{LP}\mathrm{................}[\text{reflexive property of congruency]}$

$\overline{PK}\cong \overline{PM}\mathrm{.........}[\text{radii of a triangle are congruent]}$

So, by HL Postulate , $\Delta KLP\cong \Delta \text{MLP}$ .

So, $\overline{KL}\cong \overline{ML}$

Hence , $\overline{JP}$ bisects $\overline{KM}$ .

Now, $\Delta KLP\cong \Delta \text{MLP}$

So, $\angle KPL\cong \angle MPL$

Also,

$\begin{array}{l}\angle KPJ\cong \angle KPL\\ \angle MPJ\cong \angle MPL\mathrm{..................}[\text{reflexive property of congruency]}\end{array}$

So,

$\begin{array}{l}\angle KPJ\cong \angle MPJ\mathrm{...............}[\text{transitive property of congruency]}\\ \Rightarrow m\angle KPJ=m\angle MPJ\mathrm{......}[\text{definition of congruency]}\\ \Rightarrow marcKJ=marcMJ\mathrm{.....}[\text{arc mesure is equal to the relative central angle]}\\ \Rightarrow arcKJ\cong arcMJ\mathrm{...........}[\text{definition of congruency]}\end{array}$

Hence, $\overline{JP}$ bisects arc KM.

Hence proved.