# In an isosceles triangle with base AC, angle C = 50 °, angle B = 80 ° BK is the bisector of triangle

**In an isosceles triangle with base AC, angle C = 50 °, angle B = 80 ° BK is the bisector of triangle ABC. Determine the angles of triangle ABK.**

You are given an isosceles triangle ABC.

Moreover, AC is the base of this triangle.

Now let’s determine the angles of the triangle ABK.

1.) By the condition of the problem, an isosceles triangle ABC is given.

By the property of an isosceles triangle, the angles of this triangle at its base are equal.

Thus, the angle CAB = BCA = 50 °.

The CAB angle by construction is equal to the KAB angle.

That is, we will write for the angle KAB.

KAB = 50 °.

2.) By assumption, the bisector BK is drawn to the base AC.

It divides the corner ABC in half.

Now let’s write for the angle ABC.

ABC = ABK + KBC.

Let’s find the value of the angle ABK.

Since the angle ABK is equal to the angle KBC, it is possible to write.

ABK = ABC: 2.

ABC = 80 °.

Substitute this value for the angle ABC into the formula.

ABK = 80 °: 2.

ABK = 40 °.

3.) In an isosceles triangle, the bisector drawn to the base of this triangle is also the height.

At the same time, the height of the triangle is a perpendicular drawn from the apex of the triangle to the opposite side.

That is, the bisector BK is perpendicular to the AC side of triangle ABC.

Thus, for the angle AKB of the triangle ABK we can write.

AKB = 90 °.

Answer: KAB = 50 °; ABK = 40 °; AKB = 90 °.