Question

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
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Answer 1

Step-by-step explanation:

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Answer 2

$\huge\star{\purple{\underline{\mathfrak{Explanation:}}}}$

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We have to prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

We will draw a quadrilateral ABCD which will circumscribe a circle with its centre O in a way that it touches the circle at point P, Q, R, and S.

Then,after joining the vertices of ABCD we get the figure (see in attachment shown above)

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Now, consider the triangles OAP and OAS,

AP = AS (They are the tangents from the same point A)

OA = OA (It is the common side)

OP = OS (They are the radii of the circle)

So, by SSS congruency △ OAP ≅ △ OAS

So, ∠POA = ∠AOS

Which implies that ∠1 = ∠8

Similarly other angles will be,

∠4 = ∠5

∠2 = ∠3

∠6 = ∠7

Now by adding these angles we get,

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360°

Now by rearranging,

⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360°

⇒ 2∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360°

Taking 2 as common and solving we get,

(∠1 + ∠2) + (∠5 + ∠6) = 180°

Thus, ∠AOB + ∠COD = 180°

Similarly, it is proved that ∠ BOC + ∠ DOA = 180°

$\huge\star{\green{\underline{\underline{\mathscr{Hence,proved!}}}}}$