In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC =30°, ∠BDC= 10° and ∠CAB =70°. Find:

∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC, and ∠DBA.

∠ABC = ∠ADC = 30° [Measure of opposite angles is equal in a parallelogram]

∠BDC = 10°………….. given

∠BDA = 30° - 10° = 20°

∠DAB = 180° - 30° = 150°

∠BCD = ∠DAB = 150° [Measure of opposite angles is equal in a parallelogram]

∠DBA = ∠BDC = 10° [Alternate interior angles are equal]

In ΔDOC

∠BDC + ∠ACD + ∠DOC = 180° [Sum of all angles og a triangle is 180°]

10° + 70° + ∠DOC = 180°

∠DOC = 180°- 80°

∠DOC = 100°

∠DOC = ∠AOB = 100° [Vertically opposite angles are equal]

∠DOC + ∠AOD = 180° [Linear pair]

100° + ∠AOD = 180°

∠AOD = 180°- 100°

∠AOD = 80°

∠AOD = ∠BOC = 80° [Vertically opposite angles are equal]

∠ABC + ∠BCD = 180° [In a parallelogram sum of adjacent angles is 180°]

30° + ∠ACB + ∠ACD = 180°

30° + ∠ACB + 70° = 180°

∠ACB = 180° - 100°

∠ACB = 80°

∠ACB = ∠ACB = 80° [Alternate interior angles are equal]