Question

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Iғ ʙɪsᴇᴄᴛᴏʀ ᴏғ ᴏᴘᴘᴏsɪᴛᴇ ᴀɴɢʟᴇs ᴏғ ᴀ ᴄʏᴄʟɪᴄ ǫᴜᴀʟᴅʀɪʟᴀᴛᴇʀᴀʟ ABCD ɪɴᴛᴇʀsᴇᴄᴛ ᴛʜᴇ ᴄɪʀᴄʟᴇ ᴄɪʀᴄᴜᴍsᴄʀɪʙɪɴɢ ɪɴ ᴀᴛ ᴛʜᴇ ᴘᴏɪɴᴛs P ᴀɴᴅ Q , ᴘʀᴏᴠᴇ ᴛʜᴀᴛ PQ ɪs ᴀ ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ᴀ ᴄɪʀᴄʟᴇ.



Wᴀʀɴɪɴɢ⚠️
#Sᴘᴀᴍᴍᴇʀs sᴛᴀʏ Aᴡᴀʏ......

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Answer 1

Given:-

ABCD is a cyclic quadrilateral, bisectors of ∠A and ∠C cuts the circle at Q and P

To Prove:- PQ is diameter of this circle

Construction : Join PA and PD

Proof : To prove PQ is diameter of circle ∠QAP should be equal to 90°

Now, ABCD is a cyclic quadrilateral,

then ∠A + ∠C = 180°

1/2 <A+ 1/2 <C=90°

QAD + ∠PCD = 90°

∠PCD and ∠PAD are the angles subtended by are PD is the same segment.

∴ ∠PCD = ∠PAD …(ii)

From (i) and (ii)

∠QAD + ∠PAD = 90° [∵ ∠QAD + ∠PAD = ∠QAP] ⇒ ∠QAP = 90°

⇒ ∠QAP is angle in semicircle.

Answer 2

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∠1 = ∠ 2

{Angles in the same segment of a circle are equal}

We know that the sum of opposite angles of a cyclic quadrilateral is 180°

Therefore,

∠A + ∠C = 180°

$\frac{1}{2} \:( angle \: a) \: + \frac{1}{2} \: (angle \: b)$

$\frac{1}{2} \times 180$

∠3 + ∠2 = 90°{AP and CQ are bisectors of ∠A and ∠C respectively}

∠3 + ∠1 = 90°

= ∠PAQ = 90°

Hence, PQ is a diameter of the circle ...

#Proved...

Step-by-step explanation:

Hope it's helpful.....❤️

have a good day ahead....❤️

#Thanku:)