Question

xy and x'y' are two parallel
tangents to a circle with centre
o and another tangent AB with
point of contact ç intersecting XY at A and X'Y' at B. prove that LAOB = 90°
​

Answer 1

Answer:

Consider the problem

Let us join point O to C

In ΔOPAandΔOCA

OP=OC (Radii of the same circle)

AP=AC (Tangent from point A)

AO=AO (Common side)

ΔOPA≅ΔOCA (SSS congruence criterion)

Therefore, P↔C,A↔A,O↔O

∠POA=∠COA.........(1)

Similarly,  

∠QOB≅∠OCB

∠QOB=∠COB.........(2)

Since,POQ is a diameter of the circle, it is a straight line.

Therefore, ∠POA+∠COA+∠COB+∠QOB=180  

∘

 

So, from equation (1) and equation (2)

2∠COA+2∠COB=180  

∘

 

∠COA+∠COB=90  

∘

 

∠AOB=90  

∘

 

​  

 

Step-by-step explanation:

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!!