Question

prove: opposite angles of cyclic quadrilateral are supplementary
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Answer 1

Step-by-step explanation:

Given : A cyclic quadrilateral $$ABCD$$.

To Prove : $$\angle A + \angle C = {180}^{o}$$

$$\angle B + \angle D = {180}^{o}$$

Construction : Let $$O$$ be the centre of the circle. Join $$O$$ to $$B$$ and $$D$$. Then let the angle subtended by the minor arc and the major arc at the centre be $${x}^{o}$$ and $${y}^{o}$$ respectively.

Proof : $${ x }^{ o }=2\angle C$$ [Angle at centre theorem] ...(i)

$${ y }^{ o }=2\angle A$$ ...(ii)

Adding (i) and (ii), we get

$${ x }^{ o }+{ y }^{ o }=2\angle C+2\angle A$$ ...(iii)

But, $${ x }^{ o }+{ y }^{ o }={ 360 }^{ o }$$ ....(iv)

From (iii) and (iv), we get

$$2\angle C+2\angle A={ 360 }^{ o }$$

$$\Rightarrow$$ $$ \angle C+\angle A={ 180 }^{ o }$$

But we know that angle sum property of quadrilateral

$$\angle A+\angle B+\angle C+\angle D={ 360 }^{ o }$$

$$\angle B+\angle D+{ 180 }^{ o }={ 360 }^{ o }$$

$$\angle B+\angle D={ 180 }^{ o }$$

Hence proved.

solution

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Answer 2

Answer:

❀given:

let ABCD is cyclic quadrilateral

❀ To Prove:

∠A+ ∠C=180° and ∠B+∠D=180°

❀proof:

∠BOD=2 ∠BAD

∠BAD=$\frac{1}{2}$∠BOD

Similarly ∠BCD=$\frac{1}{2}$∠DOB

∠BAD +∠BCD=$\frac{1}{2}$∠BOD+$\frac{1}{2}$∠DOB

=$\frac{1}{2}$(∠BOD+∠DOB)

=($\frac{1}{2} )$×360°=180°

similarly ∠B+∠D=180°