Question

Prove that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

Answer 1

Sol :
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

Cyclic Quadrilateral

To prove that the sum of either pair of opposite angles of cyclic quadrilateral is $180\textdegree$.  

Let ABCD is a cyclic quadrilateral of a circle with center at O.

We have to prove that  $\angle A+\angle C=\angle B+\angle D= 180\textdegree$.

construction is done for joining  BD and CD.

As we know that angle in same segment of a circle are equal.

$\angle 5=\angle8 (regarding \ chord \ AB)\\\\\angle 2=\angle4 (regarding \ chord \ CD)\\\\\angle 1=\angle6 (regarding \ chord \ BC)\\\\\angle 7=\angle3 (regarding \ chord \ AD)$

According to angle sum  property of a quadrilateral, we know that sum of all the angles of the quadrilateral is equal to 360°.

Hence in quadrilateral ,

$\angle A +\angle B+\angle C+\angle D=360\textdegree$

$\implies\angle 1+\angle 2+\angle 3+\angle 4+\angle 7+\angle 8+\angle 5+\angle 6=360\textdegree$

$\implies(\angle 1+\angle 2+\angle 7+\angle 8)+(\angle 3+\angle 4+\angle 5+\angle 6)=360\textdegree\\\\\implies2(\angle 1+\angle 2+\angle 7+\angle 8)=360\textdegree\\\\\implies (\angle 1+\angle 2+\angle 7+\angle 8)=180\textdegree\\\\\implies (\angle 1+\angle 2)+(\angle 7+\angle 8)=180\textdegree\\\\\implies \angle BAD+\angle BCD=180\textdegree\\\\\implies \angle A+\angle C=180\textdegree$

Similarly,

$\angle B+\angle D=180\textdegree$

Hence proved, that sum of opposite angles of a triangle is 180°.