Question

proof of theorem : the sum of either pair of opposite angles of a cyclic quadrilateral is 180°

Answer 1

Here is the proof of the statement provided by you.

 

Consider a cyclic quadrilateral ABCD inscribed a circle with centre at O

In order to prove this theorem, join OB and OD



 

⇒∠BAD + ∠BCD = 180.. (1)

 

Similarly, by joining OA and OC it can be proved that

∠ABC + ∠ADC = 180 ... (2)

 

Equation (1) and equation (2) shows that the sum of opposite angles of a cyclic quadrilateral is 180 degree.

 

H


The opposite angles of a cyclic quadrilateral are supplementary.

Proof



Consider a circle, with centre O. Draw a cyclic .
The aim is to prove thatand



The opposite angles of a cyclic quadrilateral are supplementary.

Proof



Consider a circle, with centre O. Draw a cyclic quadrilateral


by reason
⇒BAD + BCD = 180� ... (1) Similarly, by joining OA and OC it can be proved that ABC + ADC = 180� ... (2) Equation (1) and equation (2) shows that the sum of opposite angles of a cyclic quadrilateral is 180 degree. Hope!

Theorem: The sum of either pair of apposite angles of a cyclic quadrilateral is 180. OR The opposite angles of a cyclic quadrilateral are supplementry. Since PQRS is a cyclic quadrilateral. PSR + PQR - 180 120 + PQR = 180PQR = 60 Since PRQ is the angle in a semi-circle.PRQ = 90 Now, in PQR, we haveQPR + PRQ + PQR = 180QPR+ 90+ 60 = 180QPR = 30PLEASE THUMBS UP

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Answer 2

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°