diagonals AC and BC of parallelogram ABCD intersect at O. if angle BOC = 90 and angle BDC = 50, find angle OAB
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given that BOC=90°
similarly BOA = 90° linear pair
BDC= 50°
ABD=50° alternate interior angles
therefore in ∆AOB
BOA= 90°
ABO=50°
Angle sum property
OAB = 40°
similarly BOA = 90° linear pair
BDC= 50°
ABD=50° alternate interior angles
therefore in ∆AOB
BOA= 90°
ABO=50°
Angle sum property
OAB = 40°
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