Chapter 10.3, Problem 26PPS

To determine

Prove :

$\overline{XZ}\cong \overline{YZ},arcXB\cong arcYB$

Explanation of Solution

Given :

In $\odot C$ , $\overline{AB}\perp \overline{XY}$

Calculation:

Given: In $\odot C$ , $\overline{AB}\perp \overline{XY}$

Consider $\Delta XZC\text{ and }\Delta YZC$

$\angle XZC\text{ and }\angle YZC$ are right angles.

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

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

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

So, by HL Postulate , $\Delta XZC\cong \Delta YZC$ .

So, $\overline{XZ}\cong \overline{YZ}$

Now, $\Delta XZC\cong \Delta YZC$

So, $\angle XCZ\cong \angle YCZ$

Also,

$\begin{array}{l}\angle XCZ\cong \angle XCB\\ \angle YCZ\cong \angle YCB\mathrm{..................}[\text{reflexive property of congruency]}\end{array}$

So,

$\begin{array}{l}\angle XCB\cong \angle YCB\mathrm{...............}[\text{transitive property of congruency]}\\ \Rightarrow m\angle XCB=m\angle YCB\mathrm{......}[\text{definition of congruency]}\\ \Rightarrow marcXB=marcYB\mathrm{.....}[\text{arc mesure is equal to the relative central angle]}\\ \Rightarrow arcXB\cong arcYB\mathrm{...........}[\text{definition of congruency]}\end{array}$

Hence proved.