a point O is inside a rhombus ABCD such that OB equals to OD prove that AO and CO are in the same straight line
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AD = AB as the sides of rhombus are equal
AO = AO this is the common side
DO = BO as given
so,
ΔAOD congurent to ΔAOB
and; ∠AOD = ∠BOA
in a same way
∠COD = ∠COB
summing up the angles of rhombus account to 360 so
∠AOD + ∠BOA + ∠COD + ∠COB = 360
∠AOD + ∠AOD + ∠COD + ∠COD = 360
2∠AOD + 2∠COB = 360
∠AOD + ∠COB = 360/2
∠AOD + ∠COB = 180
so AO and Co is a straight line
AO = AO this is the common side
DO = BO as given
so,
ΔAOD congurent to ΔAOB
and; ∠AOD = ∠BOA
in a same way
∠COD = ∠COB
summing up the angles of rhombus account to 360 so
∠AOD + ∠BOA + ∠COD + ∠COB = 360
∠AOD + ∠AOD + ∠COD + ∠COD = 360
2∠AOD + 2∠COB = 360
∠AOD + ∠COB = 360/2
∠AOD + ∠COB = 180
so AO and Co is a straight line
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