In triangle ABC, angle B is 36, AB = BC, AD is the bisector. Prove that triangle is ABD-isosceles.

1. The sum of the angles of triangle ABC is 180 °:

∠A + ∠B + ∠C = 180 °, hence:
∠A + ∠C = 180 ° – ∠B;
∠A + ∠C = 180 ° – 36 ° = 144 °. (1)
2. In an isosceles triangle ABC, the angles at the base AC are:

∠A = ∠C;

From equation (1) we get:

∠A + ∠A = 144 °;
2∠A = 144 °;
∠A = 144 °: 2;
∠A = 72 °.
3. The bisector AD divides the angle A in half:

∠BAD = ∠CAD = 1/2 * ∠A = 1/2 * 72 ° = 36 °.

4. In triangle ABD, two angles are equal:

∠BAD = ∠B = 36 °,

therefore, it is isosceles, as required.