prove that the perimeter of a triangle is greater than sum of its three medians
Answers
1.)Prove that the perimeter of a triangle is greater than the sum of its 3 medians. ... Show that the sum of three altitudes of a triangle is less than sum of three sides of triangle . 3.) Prove that any two sides of triangle are together greater than twice the median drawn to the third side
Answer:
GIVEN :
A ∆ABC in which AD, BE and CF are its medians.
TO PROVE :
AB + BC + AC > AD + BE + CF
PROOF :
We know that the sum of any two sides of a triangle is greater than twice the median bisecting the third side. Therefore, AD is the median bisecting BC
=> AB + AC > 2AD (1)
• BE is the median bisecting AC
=> AB + BC > 2BE (2)
• And, CF is the median bisecting AB
=> BC + AC > 2CF (3)
Adding (i), (ii), (iii) we get,
( AB + AC ) + ( AB + BC ) + ( BC + AC ) >
2. AD + 2. BE + 2. CF
=> 2 ( AB + BC + AC ) > 2 ( AD + BE + CF )
=> AB + BC + AC > AD + BE + CF
Hence Proved !