AM is a median of ∆ABC. Prove that (AB+BC+CA) is greater than 2AM.
Please answer fast!
I will mark as brainliest!
No irrelevant answers plz.
Share
AM is a median of ∆ABC. Prove that (AB+BC+CA) is greater than 2AM.
Please answer fast!
I will mark as brainliest!
No irrelevant answers plz.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
ᴀᴍ ɪs ᴀ ᴍᴇᴅɪᴀɴ. sᴏ, ʙᴍ = ᴄᴍ
ᴄᴏɴsᴛʀᴜᴄᴛɪᴏɴ: ᴇxᴛᴇɴᴅ ᴀᴍ ᴛᴏ ᴅ, sᴜᴄʜ ᴛʜᴀᴛ ᴀᴍ= ᴍᴅ
=> ᴀʙᴅᴄ ɪs ᴀ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ ( ᴀs ᴅɪᴀɢᴏɴᴀʟs ᴀʀᴇ ʙɪsᴇᴄᴛɪɴɢ ᴇᴀᴄʜ ᴏᴛʜᴇʀ)
sɪɴᴄᴇ, ᴀʙ + ʙᴍ > ᴀᴍ……………..(1) ( ᴛʜᴇ sᴜᴍ ᴏғ 2 sɪᴅᴇs ᴏғ ᴀ ᴛʀɪᴀɴɢʟᴇ > ᴛʜɪʀᴅ sɪᴅᴇ)
& ʙᴅ + ʙᴍ > ᴍᴅ ………….(2) ( ᴛʜᴇ sᴀᴍᴇ ʀᴇᴀsᴏɴ)
ɴᴏᴡ, ʙʏ ᴀᴅᴅɪɴɢ (1) & (2)
ᴡᴇ ɢᴇᴛ, ᴀʙ + ʙᴅ + 2 ʙᴍ > ᴀᴍ + ᴍᴅ ………(3)
ʙᴜᴛ ʙᴅ = ᴀᴄ ( ᴏᴘᴘᴏsɪᴛᴇ sɪᴅᴇs ᴏғ ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ)
& 2ʙᴍ = ʙᴄ
& ᴀᴍ = ᴍᴅ
sᴏ, ᴇǫ (3) ʙᴇᴄᴏᴍᴇs
ᴀʙ + ᴀᴄ + ʙᴄ = 2ᴀᴍ
[ ᴘʀᴏᴠᴇᴅ]
Step-by-step explanation:
ʜᴏᴘᴇ ɪᴛ ʜᴇʟᴘs ʏᴏᴜ ❣️ ғᴏʟʟᴏᴡ ᴍᴇ
Verified answer
Lesson:- Lines and Angles in class 9 th , chapter 6 and excercise 6.4
As we know that the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore,
In △ABM
AB+BM>AM.....(i)
In △AMC
AC+MC>AM.....(ii)
Adding eq^n
(i)&(ii), we have
(AB+BM)+(AC+MC)>AM+AM
⇒AB+(BM+MC)+AC>2AM
⇒AB+BC+AC>2AB
Hence AB+BC+AC>2AB