Angles (2 Isosceles Triangles)

First

The figure below is not drawn to scale. ABC is an isosceles triangle where BA = BC. Given that D is the midpoint of BC and AC is 1/2 of BA, find ∠ADB.

Method 1

∠DCB = (180 - 20) / 2 = 80 (base angles of isosceles triangle ABC)

∠ADC = (180 - 80) / 2 = 50 (base angles of isosceles triangle ACD)

∠ADB = 180 - 50 = 130 (adjacent angles on straight line BDC)