In an isosceles triangle ABC with base BC, the bisector BL = 2, and the height AD = 1. Find the angle ABC.

Let’s carry out DE || BL. Since AD ​​is the height of an isosceles triangle, therefore AD is the median and bisector.

Then we get that BD = CD, and therefore DE is the middle line of the triangle BLC.

DE = 1/2 * BL = 1/2 * 2 = 1

Therefore AD = ED = 1, from which we obtain that the triangle AED is isosceles.

Let the angles DAE = AED = x, and the angles ABL = CBL = y.

EDC = LBC = y as corresponding for BL || ED and secant BC.

Angles: BAD + ABD = x + 2 * y = 90 ° – this follows from the rule about the sum of the acute angles of a right triangle.

Angles: AED = ЕDC + ЕCD: x = y + 2 * y – this equality follows from the theorem about the outer angle of a triangle (the outer angle is equal to the sum of two not adjacent to them).

x = 3 * y;

x + 2 * y = 3 * y + 2 * y = 5 * y = 90 °;

y = 18 °;

ABC = 2 * y = 36 °.

Answer: 36 °