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

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by prat23

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by prat23

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AB^2+AC^2=2BD^2+2AD^2

=2*(I/2 BC)^2+2AD^2

=1/2BC^2+2AD^2

THEREFORE 2AB^2+2AD^2=BC^2+4AD {1}

SIMILARLY

2 AB^2 +2BC^2=AC^2+4BE^2 {2}

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

ADDING 1 ,2 AND 3

4AB^2+4BC^2+4AC^2=AB^2+BC^2+4AD^2+4BE^2+4CF^2

3(AB^2+BC^2+AC^2)=4(AD^2+BE^2+CF^2)

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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)**

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,

similarly we can prove that:

add (3), (4) and (5) to get:

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