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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

We can prove that the line formed by joining the vertex to the intersection point of the arcs will always bisect the angle, no matter where the intersection point is, by using triangle congruence.

Step 1 — Setting up the construction

Let us start with an angle, $\angle XOY$ . We place the compass point at O. We draw an arc that cuts the line OX at B. This arc also cuts the line OY at A. So, the distance from O to A is the same as O to B. This means OA = OB.

Next, we place the compass point at A. We draw an arc with a certain radius. Then, we place the compass point at B. We draw another arc with the exact same radius. These two arcs meet at a point, let us call it C. So, the distance from A to C is the same as B to C. This means AC = BC. We then draw a line from O through C.

Diagram 1

Step 2 — Proving the bisection

Let us look at two triangles: $\triangle OAC$ and $\triangle OBC$ . We know that side OA is equal to side OB. $OA = OB$ We also know that side AC is equal to side BC. $AC = BC$ The side OC is common to both triangles. $OC = OC$ Since all three sides of $\triangle OAC$ are equal to the three sides of $\triangle OBC$ , the triangles are congruent. This is called the SSS (Side-Side-Side) congruence rule. $\triangle OAC \cong \triangle OBC$ When two triangles are congruent, their corresponding angles are equal. So, the angle $\angle AOC$ must be equal to the angle $\angle BOC$ . $\angle AOC = \angle BOC$ This means the line OC divides the angle $\angle XOY$ into two equal parts. Therefore, OC is the angle bisector of $\angle XOY$ . The position of point C (whether it's inside or outside the angle) does not change the fact that the triangles are congruent.

Answer

Yes, the line OC will still be an angle bisector. The construction relies on creating two congruent triangles ( $\triangle OAC$ and $\triangle OBC$ ) using the SSS congruence rule. The equality of sides (OA=OB, AC=BC, OC=OC) holds true regardless of whether point C is inside or outside the original angle. Since the triangles are congruent, the corresponding angles $\angle AOC$ and $\angle BOC$ are equal, meaning OC bisects $\angle XOY$ .

Question 9

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