Question

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

Answer 1

$\huge\red{\underline{{\bf A}}}$$\huge\pink{\underline{{\bf n}}}$$\huge\green{\underline{{\bf s}}}$$\huge\purple{\underline{{\bf w}}}$$\huge\orange{\underline{{\bf e}}}$$\huge\blue{\underline{{\bf r}}}$

ɢɪᴠᴇɴ : ᴀ ᴄɪʀᴄʟᴇ ᴡɪᴛʜ ᴄᴇɴᴛʀᴇ ᴏ ᴛᴏᴜᴄʜᴇs ᴛʜᴇ sɪᴅᴇs ᴀʙ, ʙᴄ, ᴄᴅ ᴀɴᴅ ᴅᴀ ᴏғ ᴀ ϙᴜᴀᴅʀɪʟᴀᴛᴇʀᴀʟ ᴀʙᴄᴅ ᴀᴛ ᴛʜᴇ ᴘᴏɪɴᴛs ᴘ, ϙ, ʀ ᴀɴᴅ s ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ.

ᴛᴏ ᴘʀᴏᴠᴇ : ∠ᴀᴏʙ + ∠ᴄᴏᴅ = 180°

∠ᴀᴏᴅ + ∠ʙᴏᴄ = 180°

ɢɪᴠᴇɴ : ᴀ ᴄɪʀᴄʟᴇ ᴡɪᴛʜ ᴄᴇɴᴛʀᴇ ᴏ ᴛᴏᴜᴄʜᴇs ᴛʜᴇ sɪᴅᴇs ᴀʙ, ʙᴄ, ᴄᴅ ᴀɴᴅ ᴅᴀ ᴏғ

ᴄᴏɴsᴛ. : ᴊᴏɪɴ ᴏᴘ, ᴏϙ, ᴏʀ ᴀɴᴅ ᴏs.

ᴘʀᴏᴏғ : sɪɴᴄᴇ, ᴛʜᴇ ᴛᴡᴏ ᴛᴀɴɢᴇɴᴛs ᴅʀᴀᴡɴ ғʀᴏᴍ ᴀɴ ᴇxᴛᴇʀɴᴀʟ ᴘᴏɪɴᴛ ᴛᴏ ᴀ ᴄɪʀᴄʟᴇ sᴜʙᴛᴇɴᴅ ᴇϙᴜᴀʟ ᴀɴɢʟᴇs ᴀᴛ ᴛʜᴇ ᴄᴇɴᴛʀᴇ.

∴ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8

sɪɴᴄᴇ sᴜᴍ ᴏғ ᴀʟʟ ᴛʜᴇ ᴀɴɢʟᴇs sᴜʙᴛᴇɴᴅᴇᴅ ᴀᴛ ᴀ ᴘᴏɪɴᴛ ɪs 360°.

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8

= 360°

⇒2 ∠2 + 2 ∠3 + 2 ∠6 + 2 ∠7) = 360°

⇒ 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°

⇒ ∠2 + ∠3 + ∠6 + ∠7) = 180°

⇒ (∠6 + ∠7) + (∠2 + ∠3) = 180°

⇒ ∠ᴀᴏʙ + ∠ᴄᴏᴅ = 180°

sɪᴍɪʟᴀʀʟʏ, ᴡᴇ ᴄᴀɴ ᴘʀᴏᴠᴇ ∠ᴀᴏᴅ + ∠ʙᴏᴄ = 180°

ɱαɾҡ αร ɓɾαเɳℓเεรƭ⭐

Answer 2

AnswEr:-

GIVEN :-

⇒ ABCD is a quadrilateral and it has circumscribing a circle Which has centre  O.

CONSTRUCTION ;-

⇒ Join -  AO, BO, CO, DO.

TO PROVE :-

⇒  Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

PROOF ;-

From  the given figure ,

 

⇒  ∠DAO = ∠BAO [ AB and AD are tangents in the                                                  circle]  

So , we take this angle as 1 :

         

   ⇒  ∠DAO = ∠BAO = 1

Also  in quad. ABCD ,

  ⇒ ∠ABO = ∠CBO { BA and BC are tangents }

Also , let us take this angles as 2. that is ,

 

⇒ ∠ABO = ∠CBO = 2  

let us take for vertices C and as well as D.

⇒ Sum. of angles of quadrilateral ABCD =  360° { Sum of angles of quad                                                                                   is 360°}

Therefore ,

⇒ 2 (1  + 2 + 3 + 4 )  =  360° { Sum. of angles of quad is - 360° }

         

 

⇒ 1  +  2  +  3  +  4 = 180°  

Now , in Triangle  AOB,

               

⇒ ∠BOA =  180  –   ( a + b )

                                           { Equation 1 }

Also , In triangle COD,

 

⇒ ∠COD  =  180  –  ( c + d )

                                          { Equation 2 }

From Eq. 1 and 2 we get ,

 

 ⇒ Angle  BOA + Angle  COD

                = 360 – ( a  +  b  +  c  +  d )  

             =  360°   –  180°  

                    = 180°  

So , we conclude that the line  AB and CD subtend supplementary angles at the centre  O

.

Therefore, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Hence proved

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