Question

25points for this question Given: AB ≅ BC and AO ≅ OC OK − angle bisector of ∠BOC Find: m∠AOK

Answer 1

• AB and BC are congruent (given)
• AO and OC are congruent (given)
• BO is congruent with itself (reflexivity)

Taken together, these three facts imply that triangles AOB and COB are congruent (side-side-side congruence postulate).

Now, it's not exactly clear whether A and C lie on the same line (it's possible that the figure is not drawn to scale). If they do, then both angles AOB and COB are right angles, so angle BOK has measure 45º, and so $m\angle AOK=135^\circ$.

Answer 2

By applying the theorems of an isosceles triangle and an angle bisector, m<AOK = 135 degrees.

Recall the Theorem of Isosceles Triangle:

• A triangle having two equal sides is an isosceles triangle.
• If a line segment from a vertex of the isosceles triangle meets intersects the base of the isosceles triangle at the midpoint, the line segment is therefore perpendicular to its base.
• The angles formed on both sides of the line segment that bisects the base are each 90 degrees.

We would apply the above stated to solve the problem given.

We know the following:

• AB ≅ BC (given)
• AO ≅ OC (given)
• OK bisects angle BOC (given)

We can deduce that:

• Triangle ABC is an isosceles triangle (it has two equal sides, AB and BC).
• m<AOB = 90 degrees (BO is perpendicular to AC)
• m<BOC = 90 degrees (BO is perpendicular to AC)
• m<BOK = 45 degrees (OK bisects angle BOC)

• Thus:

m<AOK = m<AOB + m<BOK

• Substitute

m<AOK = 90 + 45

m<AOK = 135 degrees.

Therefore, by applying the theorems of an isosceles triangle and an angle bisector, m<AOK = 135 degrees.

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