Q14. - Geometry

Question

a. Given: $\overline{WZ}$ is a diameter of $\odot o;\overline{OX}\parallel \overline{ZY}$

    Prove: $\stackrel{⏜}{WX}\cong \stackrel{⏜}{XY}.$

    (Hint: draw $\stackrel{⏜}{OY}.$)

Step-by-Step Solution

Answer

(a)

The final answer is

Arc length $WX\cong \text{ arc length XY.}$

Step 1. Given information:

$\overline{WZ}$ is a diameter of circle and $\overline{OX}\parallel \overline{ZY}$

Step 2. Concept used:

We use basic geometric and angle concept

Step 3. Applying the concept:

 First let us draw $\overline{OY}$ in the given figure.

Then $\overline{OY}$and $\overline{OZ}$ becomes radii of the circles $\Rightarrow \overline{OY}=\overline{OZ}$

Let $\angle WOX=x°$, since $\overline{OX} \left|\right|\overline{ZY}$ then $\angle OZY =x°$and also $\Delta OZY$ is isosceles triangles as  $\overline{OY}=\overline{OZ}$ then $\angle OYZ =x°$

From $\Delta OZY$, we have $\angle OZY+\angle ZYO+\angle YOZ={180}^0$

$\begin{array}{l}\Rightarrow x^{\circ }+x^{\circ }+\angle ZOY={180}^{\circ }\\ \Rightarrow \angle ZOY={180}^{\circ }-2x^{\circ }\end{array}$

Also from straight line  $\overline{WZ}$ , we have $\angle WOX+\angle XOY+\angle ZOY={180}^0$

$\begin{array}{c}\Rightarrow x^{\circ }+\angle XOY+{180}^{\circ }-2x^{\circ }={180}^{\circ }\\ \Rightarrow \angle XOY-x^{\circ }=180-180\\ \therefore \angle XOY=x^{\circ }\end{array}$

 

Thus, $\angle WOX=x°=\angle XOY$

$\Rightarrow$ Arcs made by same angle in a circle are congruent.

Hence, Arc length $WX \cong$arc length$XY$.