Question

In Fig, XY and X′Y′ are two parallel tangents to a circle with centre O and another

tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB

= 90°.

​

Answer 1

90=AOL

XY=X'Y'

HENCE AOB=90

HENCE PROVED

Answer 2

Answer:

• Qᴜᴇꜱᴛɪᴏɴ :

xʏ ᴀɴᴅ x'ʏ' ᴀʀᴇ ᴛᴡᴏ ᴘᴀʀᴀʟʟᴇʟ ᴛᴀɴɢᴇɴᴛꜱ ᴛᴏ ᴀ ᴄɪʀᴄʟᴇ ᴡɪᴛʜ ᴄᴇɴᴛʀᴇ ᴏ ᴀɴᴅ ᴀɴᴏᴛʜᴇʀ ᴛᴀɴɢᴇɴᴛ ᴀʙ ᴡɪᴛʜ ᴘᴏɪɴᴛ ᴏꜰ ᴄᴏɴᴛᴀᴄᴛ ᴄ ɪɴᴛᴇʀꜱᴇᴄᴛɪɴɢ xʏ ᴀᴛ ᴀ ᴀɴᴅ x'ʏ' ᴀᴛ ʙ. ᴘʀᴏᴠᴇ ᴛʜᴀᴛ ∠ᴀᴏʙ = 90°.

• ᴀɴꜱᴡᴇʀ :

• ɢɪᴠᴇɴ:

ᴡᴇ ʜᴀᴠᴇ ʙᴇᴇɴ ɢɪᴠᴇɴ ᴛʜᴀᴛ xʏ ᴀɴᴅ x'ʏ' ᴀʀᴇ ᴛᴡᴏ ᴘᴀʀᴀʟʟᴇʟ ᴛᴀɴɢᴇɴᴛꜱ ᴛᴏ ᴀ ᴄɪʀᴄʟᴇ ᴡɪᴛʜ ᴄᴇɴᴛʀᴇ ᴏ.

ᴀɴᴅ ᴀɴᴏᴛʜᴇʀ ᴛᴀɴɢᴇɴᴛ ᴀʙ ᴡɪᴛʜ ᴘᴏɪɴᴛ ᴏꜰ ᴄᴏɴᴛᴀᴄᴛ ᴄ ɪɴᴛᴇʀꜱᴇᴄᴛɪɴɢ xʏ ᴀᴛ ᴀ ᴀɴᴅ x'ʏ' ᴀᴛ ʙ.

• ᴛᴏ ᴘʀᴏᴠᴇ:

ᴡᴇ ɴᴇᴇᴅ ᴛᴏ ᴘʀᴏᴠᴇ ᴛʜᴀᴛ ∠ᴀᴏʙ = 90°.

•ᴄᴏɴꜱᴛʀᴜᴄᴛɪᴏɴ:

ᴊᴏɪɴ ᴏᴄ

• ꜱᴏʟᴜᴛɪᴏɴ:

ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ ᴛʜᴇ ᴛᴀɴɢᴇɴᴛꜱ ᴅʀᴀᴡɴ ᴛᴏ ᴀ ᴄɪʀᴄʟᴇ ꜰʀᴏᴍ ᴀɴ ᴇxᴛᴇʀɴᴀʟ ᴘᴏɪɴᴛ ᴀʀᴇ ᴇQᴜᴀʟ.

ᴛʜᴇʀᴇꜰᴏʀᴇ, ᴀᴘ = ᴀᴄ_____(1)

ɴᴏᴡ, ɪɴ Δᴘᴀᴏ ᴀɴᴅ Δᴄᴀᴏ, ᴡᴇ ʜᴀᴠᴇ

ᴀᴏ = ᴀᴏ [ᴄᴏᴍᴍᴏɴ]

ᴏᴘ = ᴏᴄ [ʀᴀᴅɪɪ ᴏꜰ ᴛʜᴇ ꜱᴀᴍᴇ ᴄɪʀᴄʟᴇ]

ᴀᴘ = ᴀᴄ [ꜰʀᴏᴍ ᴇQᴜᴀᴛɪᴏɴ 1]

=> Δᴘᴀᴏ ≅ Δᴀᴏᴄ ʙʏ ꜱꜱꜱ ᴄᴏɴɢʀᴜᴇɴᴄʏ ᴄʀɪᴛᴇʀɪᴀ.

∴ ∠ᴘᴀᴏ = ∠ᴄᴀᴏ

=> ∠ᴘᴀᴄ = 2∠ᴄᴀᴏ_____(2)

ꜱɪᴍɪʟᴀʀʟʏ, ᴡᴇ ʜᴀᴠᴇ ∠ᴄʙQ= 2∠ᴄʙᴏ__(3)

ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ ᴛʜᴀᴛ ꜱᴜᴍ ᴏꜰ ᴀɴɢʟᴇꜱ ᴏɴ ᴛʜᴇ ꜱᴀᴍᴇ ꜱɪᴅᴇ ᴏꜰ ᴛʀᴀɴꜱᴠᴇʀꜱᴀʟ ɪꜱ 180°.

∴ ∠ᴘᴀᴄ + ∠ᴄʙQ = 180°

=> 2∠ᴄᴀᴏ + 2∠ᴄʙᴏ = 180° [ꜰʀᴏᴍ 2 & 3]

=> ∠ᴄᴀᴏ + ∠ᴄʙᴏ = 180/2 = 90°____(4)

ɴᴏᴡ, ɪɴ Δᴀᴏʙ, ᴡᴇ ʜᴀᴠᴇ

∠ʙᴀᴏ + ∠ᴀʙᴏ + ∠ᴀᴏʙ = 180° [ᴀɴɢʟᴇ ꜱᴜᴍ ᴘʀᴏᴘᴇʀᴛʏ ᴏꜰ ᴀ ᴛʀɪᴀɴɢʟᴇ]

ꜰʀᴏᴍ ᴇQᴜᴀᴛɪᴏɴ 4, ᴡᴇ ʜᴀᴠᴇ

∠ᴄᴀᴏ + ∠ᴄʙᴏ + ∠ᴀᴏʙ = 180°

=> 90° + ∠ᴀᴏʙ = 180°

=> ∠ᴀᴏʙ = 180° - 90°

=> ∠ᴀᴏʙ = 90°

ᴛʜᴇʀᴇꜰᴏʀᴇ, ∠ᴀᴏʙ = 90°.

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ!!