Question

Q- ᴡʀɪᴛᴇ ᴛʜᴇ ᴘʀᴏᴏꜰ ɢɪᴠᴇɴ ᴡɪᴛʜ ᴛʜᴇ ᴛʜᴇᴏʀᴇᴍ ᴏꜰ ᴄʏᴄʟɪᴄ Qᴜᴀᴅʀɪʟᴀᴛᴇʀᴀʟ

ᴛʜᴇᴏʀᴇᴍ: ᴏᴘᴘᴏꜱɪᴛᴇ ᴀɴɢᴇʟꜱ ᴏꜰ ᴀ ᴄʏᴄʟɪᴄ Qᴜᴀᴅʀɪʟᴀᴛᴇʀᴀʟ ᴀʀᴇ ꜱᴜᴘᴘʟᴇᴍᴇɴᴛᴀʀʏ​

Answer 1

Explanation:

Theorem: Prove the opposite angles of a Cyclic Quadrilateral is Supplementary i.e have the sum of 180°.

Given: Cyclic Quadrilateral PQRS for a Circle with centre O.

To Prove: <P + <Q =180°

<Q + <S =180°

Construction: Join OS and OQ.

Proof: Arc QPS subtends <SOQ at the centre and <SRQ at the remaining part of the circle.

Therefore, <SOQ = 2<SRQ.

Major arc SRQ subtends reflex angle SOQ at the centre and <SOQ on the remaining part of the circle.

Therefore, Reflex <SOQ= 2<SPR

2x + 2y = 360°.

Therefore, x+y = 180°.

Hence, <P + <R = 180°.

Now in Quadrilateral PQRS,

<P + <Q + <R + <S = 360° [Angle Sum Property]

(<P + <R) + <Q + <S = 360°

i.e. 180° + <Q + <S = 360°

Therefore, <Q + <S = 360° - 180°

= 180°

Hence, <Q + <S = 180°.

\underline{Thus\;Proved.}

[Sorry if the figure's a bit messy.. I drew it in a hurry.]

Hope\;this\;Helps:)

Mark me as brainliest please

Answer 2

Theorem: Prove the opposite angles of a Cyclic Quadrilateral is Supplementary i.e have the sum of 180°.

Given: Cyclic Quadrilateral PQRS for a Circle with centre O.

$\huge\sf\red {To \: prove\:}$: <P + <Q =180°

» <Q + <S =180°

$\huge\sf\green{Construction\:}$

__________________________

Join OS and OQ.

Proof: Arc QPS subtends <SOQ at the centre

And;

<SRQ at the remaining part of the circle.

Therefore, <SOQ = 2<SRQ.

Major arc SRQ subtends reflex angle SOQ at the centre and <SOQ on the remaining part of the circle.

Therefore, Reflex <SOQ= 2<SPR

2x + 2y = 360°.

Therefore, x+y = 180°.

Hence, <P + <R = 180°.

Now in Quadrilateral PQRS,

<P + <Q + <R + <S = 360° [Angle Sum Property]

(<P + <R) + <Q + <S = 360°

i.e. 180° + <Q + <S = 360°

Therefore, <Q + <S = 360° - 180°

= 180°

$\huge\sf\blue{☆Hence, \: <Q\: + \:<S \:= 180°\:}$

$\huge\sf\purple{☆Hope\: this \:will\:be \:helpful\:to\: you\:}$

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Itzsidhu193

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