Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 45° and ∠CDB =55°, then ∠BOD =
AC ‖ BD
∠CAD = 45o
∠CDB = 55o
∠2 = ∠CAD (Alternate angle)
∠2 = 45o
In
∠BOD + ∠2 + ∠CDB = 180o
∠BOD + 45o + 55o = 180o
∠BOD + 100o = 180o
∠BOD = 180o – 100o
= 80o