prove that the sum of opposite angles of a cyclic quadrilateral is 180 degree
Answers
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°
Let a cyclic quadrilateral ABCD
To prove ∠A+∠C =180° and ∠B+∠D=180°
Construction Join AC and BD
Proof We Have
- ∠ACB=∠ADB [angles on the same segment]
- and ∠BAC=∠BDC [angles on same segment]
∴ ∠ACB+∠BAC=∠ADC
∠ACB+ ∠BAC+ ∠ABC= ∠ADC+ ∠ABC
[adding ∠ABC on both sides]
∠ADC+ ∠ABC=180° [the sum of angle of triangle ABC is 180°]
∠B+ ∠D=180° ...(1)
Now, ∠A+ ∠B+ ∠C+ ∠D=360° [ Sum of angles of quadrilateral is 360°]
∠A+ ∠C=360°-( ∠B+ ∠D)=360°-180°=180° [using 1]
Hence, ∠A+ ∠C=180° and ∠B+ ∠D=180°
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