In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line still be an angle bisector? Explore this through construction, and then justify your answer.

Understand the Question In standard angle bisection, arcs of equal radius are drawn in the interior of the angle to intersect at a point. Drawing the intersecting arcs on the other side creates two triangles, and , that share the common side . By showing that using the SSS (Side-Side-Side) congruence criterion, we can prove that , meaning the ray still bisects . Step 1 · Set Up the Construction Let the given angle be with vertex . 1. With as centre, draw an arc intersecting at and at , so: 2. With and as centres and with equal radii, draw arcs intersecting at point on the other side, so: 3. Join . Step 2 · Prove Bisection Using Triangle Congruence In and : By SSS congruence criterion: Since corresponding parts of congruent triangles are equal (CPCT): Therefore, the line bisects . Answer > Yes, the line will still be an angle bisector because by the SSS congruence criterion, which gives . Common Mistakes Assuming Arcs Must Be in the Interior: Thinking the intersection point must lie inside the angle. The geometric congruence holds regardless of which side the arcs intersect. Incorrect Congruence Rule: Using the SAS rule when angles are unknown. Since all three pairs of corresponding sides (, , ) are known to be equal, the SSS criterion must be used.

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Question 9

In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line $OC$ still be an angle bisector? Explore this through construction, and then justify your answer.

Solution

Understand the Question

In standard angle bisection, arcs of equal radius are drawn in the interior of the angle to intersect at a point.
Drawing the intersecting arcs on the other side creates two triangles, $\triangle OAC$ and $\triangle OBC$ , that share the common side $OC$ .
By showing that $\triangle OAC \cong \triangle OBC$ using the SSS (Side-Side-Side) congruence criterion, we can prove that $\angle AOC = \angle BOC$ , meaning the ray $OC$ still bisects $\angle XOY$ .

Step 1 · Set Up the Construction

Let the given angle be $\angle XOY$ with vertex $O$ . Diagram 1

With $O$ as centre, draw an arc intersecting $OY$ at $A$ and $OX$ at $B$ , so: $OA = OB$
With $A$ and $B$ as centres and with equal radii, draw arcs intersecting at point $C$ on the other side, so: $AC = BC$
Join $OC$ .

Step 2 · Prove Bisection Using Triangle Congruence

In $\triangle OAC$ and $\triangle OBC$ :

\begin{aligned} OA &= OB \quad (\text{Radii of the same arc}) \\[0.6em] AC &= BC \quad (\text{Arcs of equal radii}) \\[0.6em] OC &= OC \quad (\text{Common side}) \end{aligned}

By SSS congruence criterion: $\triangle OAC \cong \triangle OBC$

Since corresponding parts of congruent triangles are equal (CPCT): $\angle AOC = \angle BOC$

Therefore, the line $OC$ bisects $\angle XOY$ .

Answer

Yes, the line $OC$ will still be an angle bisector because $\triangle OAC \cong \triangle OBC$ by the SSS congruence criterion, which gives $\angle AOC = \angle BOC$ .

Common Mistakes

Assuming Arcs Must Be in the Interior: Thinking the intersection point $C$ must lie inside the angle. The geometric congruence holds regardless of which side the arcs intersect.
Incorrect Congruence Rule: Using the SAS rule when angles are unknown. Since all three pairs of corresponding sides ( $OA = OB$ , $AC = BC$ , $OC = OC$ ) are known to be equal, the SSS criterion must be used.

Question 9

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