# Given: triangle ABC-equilibration, AC-base, angle A = 50 deg, AD-bisector of angle A, CP-bisector of angle C

**Given: triangle ABC-equilibration, AC-base, angle A = 50 deg, AD-bisector of angle A, CP-bisector of angle C, CP and AD intersect at point O. Find: angle AOC-?**

1.) First, find the value of the angle C in triangle ABC.

Since triangle ABC is isosceles, the angles at its base are equal.

By the condition AC is the base.

This means that angles A and C are equal.

A = C = 50 °.

2.) Now consider the triangle AOC.

By construction, AD and CP are the bisectors of the angles A and C of triangle ABC, respectively.

According to the rule, the bisector divides the angle in half.

Therefore, you can write down.

DAC = OAC = PCA = OCA = BAC: 2 = PAC: 2 = BCA: 2 = DCA: 2.

PAC = BAC = DCA = BCA = A = C = 50 °.

OAC = OCA = DCA: 2 = 50 ° = 25 °.

3.) Since, according to the rule, the sum of the values of all the angles of any triangle is equal to 180 °, then we write down for the angles of the triangle AOC.

A + C + O = 180 °.

In triangle AOC, angles A and C are equal.

A = C = 25 °.

2 × 25 ° + O = 180 °.

50 ° + O = 180 °.

O = 180 ° – 50 °.

O = 130 °.

AOC = 0 = 130 °.

Answer: AOC = 130 °.