In an isosceles triangle ABC, AD is the bisector of the angle at base AC at CAD = 25 degrees. Find angle B.

Since, by condition, BP is the bisector of the BAC angle, and the SBP angle = 250, then the BAC angle = 2 * 25 = 50.

In an isosceles triangle ABC, the coals at its base are equal, then the angle ACB = BAC = 50.

The sum of the inner angles of the triangle is 180, then the angle ABC = (180 – BAC – ACB) = (180 – 50 – 50) = 80.

Answer: Angle ABC is 80.

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