Justify why in Fig. 6.4 is the perpendicular bisector.

Understand the Question A perpendicular bisector of a line segment is a line that intersects it at a right angle () and divides it into two equal halves. To justify that is the perpendicular bisector of , where is their intersection point, we must prove two things: 1. Bisector: 2. Perpendicular: This is proved by applying triangle congruence criteria (SSS followed by SAS). Step 1 · Prove Congruence of and From the construction using equal lengths from and : In and : By SSS congruence criterion: Therefore, corresponding angles are equal: Step 2 · Prove Congruence of and Let be the point of intersection of and . In and : By SAS congruence criterion: Step 3 · Show that Bisects Perpendicularly Since : 1. Corresponding sides are equal: Thus, bisects . 2. Corresponding angles are equal: Since is a straight line: Thus, , which means is perpendicular to . Answer > Since and , is the perpendicular bisector of . Common Mistakes Skipping the First Congruence Step: Jumping directly to without first proving using the larger triangles and . Incomplete Perpendicular Bisector Justification: Proving that (bisector) but forgetting to prove that the angle of intersection is (perpendicular), or vice versa.

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Question 5

Justify why $\text{AB}$ in Fig. 6.4 is the perpendicular bisector.

Solution

Understand the Question

A perpendicular bisector of a line segment is a line that intersects it at a right angle ( $90^\circ$ ) and divides it into two equal halves.
To justify that $\text{AB}$ $AB$ is the perpendicular bisector of $\text{XY}$ $XY$ , where $\text{O}$ $O$ is their intersection point, we must prove two things:
1. Bisector: $\text{OX} = \text{OY}$
2. Perpendicular: $\angle \text{XOA} = \angle \text{YOA} = 90^\circ$
This is proved by applying triangle congruence criteria (SSS followed by SAS).

Step 1 · Prove Congruence of $\triangle \text{AXB}$ and $\triangle \text{AYB}$

From the construction using equal lengths from $\text{X}$ and $\text{Y}$ : $\text{AX} = \text{AY} = \text{BX} = \text{BY}$ Diagram 1

In $\triangle \text{AXB}$ and $\triangle \text{AYB}$ :

\begin{aligned} \text{AX} &= \text{AY} \quad (\text{equal lengths}) \\ \text{BX} &= \text{BY} \quad (\text{equal lengths}) \\ \text{AB} &= \text{AB} \quad (\text{common side}) \end{aligned}

By SSS congruence criterion: $\triangle \text{AXB} \cong \triangle \text{AYB}$

Therefore, corresponding angles are equal: $\angle \text{XAB} = \angle \text{YAB} \implies \angle \text{XAO} = \angle \text{YAO}$

Step 2 · Prove Congruence of $\triangle \text{AXO}$ and $\triangle \text{AYO}$

Let $\text{O}$ be the point of intersection of $\text{AB}$ and $\text{XY}$ .

In $\triangle \text{AXO}$ and $\triangle \text{AYO}$ :

\begin{aligned} \text{AX} &= \text{AY} \quad (\text{given}) \\ \angle \text{XAO} &= \angle \text{YAO} \quad (\text{proved above}) \\ \text{AO} &= \text{AO} \quad (\text{common side}) \end{aligned}

By SAS congruence criterion: $\triangle \text{AXO} \cong \triangle \text{AYO}$

Step 3 · Show that $\text{AB}$ Bisects $\text{XY}$ Perpendicularly

Since $\triangle \text{AXO} \cong \triangle \text{AYO}$ :

Corresponding sides are equal: $\text{OX} = \text{OY}$ Thus, $\text{AB}$ bisects $\text{XY}$ .
Corresponding angles are equal: $\angle \text{XOA} = \angle \text{YOA}$

Since $\text{XOY}$ is a straight line:

\begin{aligned} \angle \text{XOA} + \angle \text{YOA} &= 180^\circ \\ 2 \angle \text{XOA} &= 180^\circ \\ \angle \text{XOA} &= \dfrac{180^\circ}{2} = 90^\circ \end{aligned}

Thus, $\angle \text{XOA} = \angle \text{YOA} = 90^\circ$ , which means $\text{AB}$ is perpendicular to $\text{XY}$ .

Answer

Since $\text{OX} = \text{OY}$ and $\angle \text{XOA} = \angle \text{YOA} = 90^\circ$ , $\text{AB}$ is the perpendicular bisector of $\text{XY}$ .

Common Mistakes

Skipping the First Congruence Step: Jumping directly to $\triangle \text{AXO} \cong \triangle \text{AYO}$ without first proving $\angle \text{XAO} = \angle \text{YAO}$ using the larger triangles $\triangle \text{AXB}$ and $\triangle \text{AYB}$ .
Incomplete Perpendicular Bisector Justification: Proving that $\text{OX} = \text{OY}$ (bisector) but forgetting to prove that the angle of intersection is $90^\circ$ (perpendicular), or vice versa.

Question 5

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