How we prove that in a triangle, any two sides of atriagle is greater than the median of third side?
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Hello Mate!
Given : In ∆ABC, AD is median.
To prove : AB + AC > AD
To construct : Draw DE = AD and join CE.
Proof : In ∆ABD and ∆ CED we have,
BD = CE [ Since AD was median ]
AD = DE [ Through construction ]
< ADB = < CDE [ Vertically opp. angles ]
Hence both triangle are congruent by SAS congruency.
AB = CE [ c.p.c.t ]
Now, in ∆ACE we have,
AC + CE > AE
AC + AB > AD + DE
AB + AC > 2AD.
ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ
Have great future ahead!
Given : In ∆ABC, AD is median.
To prove : AB + AC > AD
To construct : Draw DE = AD and join CE.
Proof : In ∆ABD and ∆ CED we have,
BD = CE [ Since AD was median ]
AD = DE [ Through construction ]
< ADB = < CDE [ Vertically opp. angles ]
Hence both triangle are congruent by SAS congruency.
AB = CE [ c.p.c.t ]
Now, in ∆ACE we have,
AC + CE > AE
AC + AB > AD + DE
AB + AC > 2AD.
ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ
Have great future ahead!
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SmãrtyMohït:
nice
Thanks dear
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