if the sum of a pair of opposite angles of a quadrilateral is 180,then show that the quadrilateral is cyclic

Theorem. Sum of the opposite angles of a cyclic quadrilateral is 180°.
Given : A cyclic quadrilateral ABCD
To prove : ∠BAD + ∠BCD = ∠ABC + ∠ADC = 180°
Construction : Draw AC and DB
Proof : ∠ACB = ∠ADB
and ∠BAC = ∠BDC
[Angles in the same segment]
∴ ∠ACB + ∠BAC = ∠ADB + ∠BDC = ∠ADC
Adding ∠ABC on both the sides, we get
∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC
But ∠ACB + ∠BAC + ∠ABC = 180° [Sum of the angles of a triangle]
∴ ∠ADC + ∠ABC = 180°
∴ ∠BAD + ∠BCD = 360° – (∠ADC + ∠ABC) = 180°.
Hence proved.
Converse of this theorem is also true.
If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.
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wrong proof 

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 ya it is

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GIVEN :       ABCD is a quadrilateral 

TO PROOF:   ANGLE A+ANGLE C=180

PROOF:   arc BCD  subtitute ANGE BOD at the center                                                                                                                                                                  and ANGLE BAD at the point A on the remaining part of the circle 

THERE FOR       ANGLE BOD =2* ANGLE BAD ~~~~~~~~~~~(1)      {THE ANGLE SUPTENTED BY AN ARC AT THE CENTER IS

                                                                                                                                           DOUBLE THE ANGE SUBTENDED BY IT AT THE                                                                                                                                                                 REMAINING PART OF THE CIRCLE}                     similarlly major arc  BAD subtentes reflex ANGLE BOD at the center 

and ANGLE BCD at the point C  onthe remaining part of the circle 

  THERE FOR     reflex ANGLE =2* ANGLE BCD ~~~~~~~~~~~~~~(2)

               ADDING  (1) AND (2)

                                   ANGLE BOD +reflex ANGLE BOD=2*ANGLE BAD +2*ANGLE BCD 

                                                              360=2{ANGLE BAD + ANGLE BCD}

                                                              360/2 =ANGLE BAD + ANGLE BCD 

  THERE FOR                                                    ANGLE A + ANGLE C=180 ~~~~~~~~~~~(A)

IN quadrilateral ABCD, ANGLE A + ANGLE B + ANGLE C + ANGLE D =360    {ANGLE SUM PROPERTY OF THE QUADRILATERAL}

                                          180+ANGLE B + ANGLE D = 360

                                                  ANGLE B + ANGLE D = 360-180

                                                             ANGLE B + ANGLE D= 180

THUS IT IS PROVED THAT THE SUM OF EIGHTHER PAIR OF OPPOSITE ANGLES OF A CYCLIC QUADRILATERAL IS 180 

 

 

 

 

 

 

 

 

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and 
 

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 If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

Proof
Image

Consider a quadrilateral   , such that   and   .

The aim is to prove that points A, B, P and Q lie on the circumference of a circle.

By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let the circle cut   (or   extended) at point R. Draw   .

(8)

  the assumption that the circle does not pass through P, must be false, and A, B, P and Q lie on the circumference of a circle and   is a cyclic quadrilateral.

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Where 8s the picture
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if sum of a pair of opposite angles of a quadrilateral is 180 then it is cyclic
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Gghhhh
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Theorem 10.12 : If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic
 
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