Chapter 10.3, Problem 29PPS

To determine

Prove that $\overline{FG}\cong \overline{JH}$ .

Explanation of Solution

Given :

LX = LY

Calculation:

Given : LX = LY

To Prove : $\overline{FG}\cong \overline{JH}$

Consider $\Delta XGL\text{ and }\Delta \text{YHL}$ :

$\overline{LG}\cong \overline{LH}\mathrm{...........}[\text{radii of a circle are congruent]}$

$\overline{LX}\perp \overline{FG}$

$\overline{LY}\perp \overline{JH}$

So, $\angle LXG\text{ and }\angle LYH$ are right angles.

$\begin{array}{l}LX=LY\\ \Rightarrow \overline{LX}\cong \overline{LY}\mathrm{...........}[\text{definition of congruency]}\end{array}$

So, by HL Postulate , the right $\Delta XGL\cong \Delta \text{YHL}$ .

Hence,

$\begin{array}{l}\overline{XG}\cong \overline{YH}\\ \Rightarrow XG=YH\mathrm{.............}[\text{definition of congruency]}\\ \Rightarrow 2(XG)=2(YH)\mathrm{........}(1)\mathrm{....}[\text{multiply each side by 2]}\end{array}$

$\overline{LX}\text{ and }\overline{LY}$ are perpendicular bisectors to the chords $\overline{FG}\text{ and }\overline{JH}$ , since $\overline{LX}\text{ and }\overline{LY}$ are contained in radii and radius that is perpendicular to the chord bisects it .

So,

$\begin{array}{l}FG=2(XG)\\ JH=2(YH)\end{array}$

Substituting in equation (1) :

$\begin{array}{l}FG=JH\\ \Rightarrow \overline{FG}\cong \overline{JH}\mathrm{..........}[\text{definiton of congruency]}\end{array}$

Hence proved.