# Prove that in a triangle the sum of the squares of the side is 4 times the sum of the squares of the median

2
by prat23

2015-09-29T19:54:23+05:30
According to appollonius theorem  the sum of the square of 2 sides  of a triangle is equal to twice the square of the median on the third side  plus half the square of the third side.

SIMILARLY
2 AB^2 +2BC^2=AC^2+4BE^2 {2}
2BC^2+2AC^2=AB^2+4CF^2 {3}

2015-09-30T00:26:25+05:30

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See diagram.

AD is the median on to BC. BE and CF are medians onto AC and AB respectively.  Draw a normal AG from A on to BC.

AB² = AG² + BG²
= AD² - DG² +  BG²
= AD² + (BG + DG) (BG - DG)
= AD² + BD (BD -  DG - DG)
= AD² + BD² -  2 * BD * DG      ---- (1)

AC² = AG² + GC²
= AD² - GD² + GC²
= AD² + (GC - GD) (GC + GD)
= AD² + DC * (2 GD + DC)
= AD² + DC² + 2 DC * GD  --- (2)

now we obtain the Apolloneous theorem, as below.  as  DC = BD = 1/2 BC,

AC² + AB² = 2 AD² + 2 BD²   = 2 AD² + 1/2 BC²   --- (3)

similarly we can prove that:
BC² + BA² = 2 BE² + 2 AE² = 2 BE² + 1/2 AC²        --- (4)
CB² + CA² = 2 CF² + 2 AF² = 2 CF² + 1/2 AB²          --- (5)

add  (3), (4) and (5) to get:

2 (AB² + AC² + BC²) = 2(AD² + BE² + CF²) + 1/2 * (AB²+ AC²+BC²)

3 (AB² + AC² + BC²)  = 4 ( AD² + BE² + CF²)

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