prove that the sum of a pair of opposite angles of a quadrilateral is 180°
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Given : Let ABCD is cyclic quadrilateral.
To prove : ∠A + ∠C = 180° and ∠B + ∠D = 180°.
Construction : join OB and OD.
Proof : ∠BOD = 2 ∠BAD
∠BAD = 1/2∠ BOD
Similarly ∠BCD = 1/2 ∠DOB
∠BAD + ∠BCD = 1/2∠BOD + 1/2 ∠DOB
=1/2(∠ BOD + ∠DOB)
= (1/2)X360° = 180°
Similarly ∠B + ∠D = 180°
To prove : ∠A + ∠C = 180° and ∠B + ∠D = 180°.
Construction : join OB and OD.
Proof : ∠BOD = 2 ∠BAD
∠BAD = 1/2∠ BOD
Similarly ∠BCD = 1/2 ∠DOB
∠BAD + ∠BCD = 1/2∠BOD + 1/2 ∠DOB
=1/2(∠ BOD + ∠DOB)
= (1/2)X360° = 180°
Similarly ∠B + ∠D = 180°
sourabhbajaj:
plz give diagram also
YES I CAN'T DRAW IN THIS APPS TRYING ...,,,,
draw on page and send pic
north side A south west B south C south east D outer side circle inner quadltrl mid point O
I think it helpsssss
thnks
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