Here is a nice trapezoid, ABCD, with AB parallel to CD. Diagonal BD is drawn, and BD = AD. ∠DCB = 110° and ∠CBD = 30°. What is ∠ADB?

- 80°
- 90°
- 100°
- 110°
- 120°

110° + 30° + B1 = 180° (co-interior angles)

B1 = 40°

∠A = 40° (isosceles triangle)

∴ D1 = 100° (= ∠ADB)

The answer is (c).