In an isosceles triangle ABC, angle B = 120 degrees, O is the intersection point of the bisectors. A circle with a radius

In an isosceles triangle ABC, angle B = 120 degrees, O is the intersection point of the bisectors. A circle with a radius of 2√3 cm is inscribed in this triangle and touches lines BC and AC at points D and E, respectively. Find the BO and the BED angle.

In a triangle BED, BE = BD as tangents drawn from one point, then BED is isosceles, then the angle BED = (180 – EBD) / 2 = 60/2 = 30.

OE is the radius of the circle drawn to the point of tangency, then the angle OEB = 90, and the triangle OEB is rectangular. BO is the bisector of the angle EBD, then the angle OBE = 60.

In a right-angled triangle OBE SinOBE = OE / OB = R / OB.

ОВ = R / Sin60 = (2 * √3) / (√3 / 2) = 4 cm.

Answer: The length of the OB segment is 4 cm, the BED angle is 30.