Question

prove that the sum of opposite angles of a cyclic quadrilateral is 180 degree​

Answer 1

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°

Answer 2

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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