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.

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

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