A kite

ABCD is a kite. Angle OBC = 20° and angle OCD = 35°. O is the intersection of diagonals. Find angle ABC, angle ADC and angle BAD.

Correct result:

x =  40 °
y =  110 °
z =  55 °

Solution:

∠OBC=20∘ ∠OCD=35∘  OBA=∠OBC=20=20∘ ABC=OBA+∠OBC=20+20=40∘  x=ABC=40=40∘
OAD=∠OCD=35=35∘ ∠ADO=90−OAD=90−35=55∘ ODC=∠ADO=55=55∘  ADC=∠ADO+ODC=55+55=110∘  y=ADC=110=110∘
BAD=OBA+OAD=20+35=55∘  z=BAD=55=55∘

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