The bisector of the outer angle at the vertex B of the triangle ABC intersects the extension of the bisector of the angle

The bisector of the outer angle at the vertex B of the triangle ABC intersects the extension of the bisector of the angle A of this triangle at the point O. The AOB angle is 80o. Find the degree measure of the angle ACB.

The main equality: the angles of the triangle ABC (<A + <B + <C) = 180 °.

Consider a triangle ABO, in which <AOB = 80 ° is known. We write the equality for the ABO triangle:

<BAO = <A / 2, since AO is the bisector of the angle BAC, which is <A
<AVO = <B + (180 – <B) / 2 = 90 ° + <B / 2.
<BOA + <BAO + <ABO = 180 °, as the sum of the angles in the ABO triangle.

Let’s make the equation:

<VOA = 80 ° = 180 ° – (<VAO + <AVO);
80 ° = 180 ° – (<A / 2 + 90 ° + <B / 2) = 90 ° – (<A / 2 + <B / 2).

Hence: (<A / 2 + <B / 2) = 10 °, (<A + <B) = 2 * 10 ° = 20 °.
Now we find the angle ACB = <C = 180 – (<A + <B) = 180 ° – 20 ° = 160 °.