Question

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

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

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⫸ Prove that Opposite Sides of Quadrilateral Circumscribing a circle subtended supplementary angles at the centre of the Circle.

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According To given statement, the fig. will be as shown in attachment, in which the quadrilateral ABCD curcumscribes the circle with centre O and its sides touches the circle at P, Q, R and S as shown.

Join OP, OQ, OR and OS

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⪼ $\angle$ AOB = $\angle$ COD = 180°

⪼ $\angle$ AOD = $\angle$ BOC = 180°

AOC and BOD are not diameters

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In $\triangle$ ROC and $\triangle$ QOC

OR = OQ (Radii)

OC = OC (Common)

CR = CQ

(Tangent Drawn to Circle from an external point are equal in length.)

So, $\sf \triangle ROC\cong \triangle QOC$. (By SSS)

$\therefore \angle 1 = \angle 2$. (By CPCT)

Similarly,

$\angle 4 = \angle 3$ (By CPCT)

$\angle 5 = \angle 6$ (By CPCT)

$\angle 8 = \angle 7$ (By CPCT)

Now,

$\sf \angle ROQ + \angle QOP + \angle POS + \angle SOR$ = 360°

(because $\angle$ O = 360°)

$\therefore \angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8$ = 360°

Now,

$:\to \angle 1 + \angle 1 + \angle 4 + \angle 4 + \angle 5 + \angle 5 + \angle 8 + \angle 8$ = 360°

Because

$\angle 1 = \angle 2$

$\angle 4 = \angle 3$

$\angle 5 = \angle 6$

$\angle 8 = \angle 7$

So,

$:\to 2\angle 1 + 2\angle 4 + 2\angle 5 + 2\angle 8$ = 360°

$:\to 2(\angle 1 + \angle 4 + \angle 5 + \angle 8)$ = 360°

$:\to \angle 1 + \angle 4 + \angle 5 + \angle 8$ = 360/2

$:\to \angle 1 + \angle 4 + \angle 5 + \angle 8$ = 180°

$:\to \angle 1 + \angle 8 + \angle 5 + \angle 4$ = 180°

So,

$:\to \angle DOC + \angle AOB$ = 180°

and $\angle AOD + \angle BOC$ = 180°

$\sf\because \angle AOD + \angle BOC + \angle DOC + \angle AOB$ = 360°

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Hence, Proved That

• $\angle$ AOB = $\angle$ COD = 180°
• $\angle$ AOD = $\angle$ BOC = 180°