Proving AO as the Angle Bisector in a Circle with Equal Chords

In the realm of geometry, particularly in the study of circles, there are several interesting theorems and properties that we can explore. One such fascinating property involves the behavior of angles and equal chords within a circle. In this article, we will delve into a detailed proof to establish that AO is the angle bisector of angle BAC in a circle where chords AB and AC are equal. Let's begin by understanding the setup and then move on to the proof.

Setup and Understanding

Consider a circle with center O. Points A, B, and C are on the circumference of the circle such that chord AB and chord AC are of equal length. Our goal is to prove that line segment AO is the angle bisector of angle BAC. To do this, we will employ several geometric properties and theorems. Let's start by drawing the figure.

Figure Explanation

When we draw the figure, the following observations can be made:

BC is the diameter of the circle because angle BAC, which is subtended by the diameter, is a right angle (by the Angle in a Semicircle Theorem). This implies that AO is a median in triangle ABC. Because O is the center of the circle, BO, AO, and CO are all radii, making them equal in length: BO AO CO.

Proof

We will now proceed with the proof step-by-step, using the above observations and geometric theorems.

Step 1: Identifying Key Angles and Sides

First, let's identify the key angles and sides in the setup:

Because AB and AC are equal chords, the angles opposite these chords at the center (angle BAO and angle CAO) will be equal. Since B, A, and C lie on the circle and O is the center, angle BAO angle CAO 45° (as the bisector of a right angle).

Since angle BAO angle CAO, we now know that in triangle AOB and triangle AOC, the corresponding angles at B and C are equal.

Step 2: Applying Congruency Criteria

Next, we will use the criteria for triangle congruency.

In triangles AOB and AOC: AO is common to both triangles. BO and CO are radii of the same circle, so BO CO. Angle BAO angle CAO (both are 45°).

This means that the two triangles AOB and AOC are congruent by the SAS (Side-Angle-Side) Congruence Criterion.

Step 3: Conclusion

Because triangles AOB and AOC are congruent, their corresponding sides are equal. Therefore, AB AC and AO is the angle bisector of angle BAC.

Alternative Proof 1: AAS Congruency

We can also prove this using the AAS (Angle-Angle-Side) Congruence Criterion in triangles ABO and ACO.

In triangles ABO and ACO: AO is common. BO and CO are radii of the same circle, so BO CO. Angle BAO angle CAO (both are 45°).

Thus, triangles ABO and ACO are congruent by the AAS criterion, and this again implies that AB AC.

Alternative Proof 2: Right Angles and Congruency

Another way to prove this is by extending AO to point D on the circumference and using the Angles in a Semicircle Theorem again.

Extend AO to point D on the circumference. Since AB and AC are equal chords, triangles ABD and ACD are isosceles with AB AC and AD AD. The angles subtended by the equal chords in the same segment are equal, so angle ABD angle ACD 90°. Thus, triangles ABD and ACD are congruent by the Right Angle-Side-Side (RAS) Congruence Criterion. Since the triangles are congruent, angle BAD angle CAD, and AO is the bisector of angle BAC.

Conclusion

Through detailed geometric analysis and the application of triangle congruency theorems, we can conclusively prove that AO is the angle bisector of angle BAC in a circle with equal chords AB and AC. The proof is robust and employs both SAS and AAS criteria, ensuring a comprehensive and thorough understanding of the geometric properties involved.