Question

AM is a median of ∆ABC. Prove that (AB+BC+CA) is greater than 2AM.
Please answer fast!
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No irrelevant answers plz.​

Answer 1

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

Answer 2

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