The diagonal BD of the trapezoid ABCD divides it into two isosceles triangles (AB = AD, CD = BD).

The diagonal BD of the trapezoid ABCD divides it into two isosceles triangles (AB = AD, CD = BD). Find the value of the angle BAD, if the value of the angle BDC is 120

Consider an isosceles triangle ∆ВDC: СD = ВD, so DBC = ВСD.

Also ВDC + DVS + ВСD = 180 °, with ВDC = 120 ° we have:

DWS + ВСD = 180 – 120 = 60 °, DWS = ВСD = 30 °.

Consider a trapezoid ABCD: the sum of the angles belonging to the lateral side is 180 °,

those. BCD + CDA = 180 °, whence CDA = 180 – 30 = 150 °.

CDA = BDC + ADB, we get ADB = CDA – BDC = 150 – 120 = 30 °.

Consider an isosceles triangle ∆ABD: AB = AD, so ABD = ADB = 30 °.

BAD = 180 – (ABD + ADB) = 180 – 60 = 120 °.

Answer: BAD = 120 °.