# 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 °. 