AD, BE and CF are medians of the triangle ABC, then prove that
3(AB2 + BC2 + CA2) = 4(AD2 + BE2 + CF2)
Answers
HELLO DEAR,
CONSTRUCTION:-
1). draw AM perpendicular BC
2). draw BN perpendicular AC
3). draw AB perpendicular CK
now,
it is given that:-
AD , BE AND CF ARE THE MEDIANS OF ∆ABC
[figure is in the attachment]
BD = DC = 1/2BC
FB = AF = 1/2AB
AE = EC = 1/2AC
IN ∆AMB AND ∆ AMC, <M = 90°
⇒AB² = BM² + AM²
⇒AB² = (DC - MD)² + AM²
[BD = DC , BM = DC - MD]
and, AC² = MC² + AM²
⇒AC² = (MD + DC)² + AM²
[BD = DC , MC = MD + DC]
NOW,
AB² + AC² = DC² + MD² - 2MD*DC + AM² + MD² + DC² + 2MD*DC + AM²
⇒AB² + AC² = 2DC² + 2MD² + 2AM²
⇒AB² + AC² = 2(1/2BC)² + 2(MD² + AM²)
⇒AB² + AC² = 1/2(BC²) + 2AD²-----------( 1 )
[ as , <M = 90°]
IN ∆ AKC AND ∆AKB, <K = 90°
⇒AC² = AK² + CK²
AND,⇒BC² = KC² + BK²
NOW,
⇒AC² + BC² = AK² + KC² + KC² + BK²
⇒AC² + BC² = (BF - KF)² + KC² + KC ² + (BF + FK)²
[BF = AF , AK = BF - FK , BK = BF + FK]
⇒AC² + BC² = BF² + FK² - 2FK*BF + KC² + KC² + BF² + FK² + 2FK*BF
⇒AC² + BC² = 2BF² + 2FK² + 2KC²
⇒AC² + BC² = 2(1/2AB)² + 2(FK² + KC²)
⇒AC² + BC² = 1/2(AB²) + 2FC²----------( 2 )
[as, <K = 90°]
similarly, IN ∆ ANB AND ∆BNC, <N = 90°
⇒AB² = AN² + BN²
AND,⇒BC² = CN² + BN²
now,
⇒AB² + BC² = AN² + CN² + 2BN²
⇒AB² + BC² = (CE - NE)² + (CE + NE)² + 2BN²
[AE = CE , AN = CE - NE, CN = CE + NE]
⇒AB² + BC² = CE² + NE² - 2CE*NE + CE² + NE² + 2CE*NE + 2BN²
⇒AB² + BC² = 2CE² + 2NE² + 2BN²
⇒AB² + BC² = 2(1/2AC)² + 2(NE² + BN²)
⇒AB² + BC² = 1/2(AC)² + 2BE²
[as , <K = 90°]
now, adding------( 1 ) , -------( 2 ) & ------( 3 )
we get,
⇒AB² + AC² + AC² + BC² + AB² + BC² = 1/2(BC)² + 2AD² + 1/2(AB)² + 2FC² + 1/2(AC)² + 2BE²
⇒2(AB² + BC² + AC²) = 1/2(AB² + BC² + AC²) + 2(AD² + FC² + BE²)
⇒2(AB² + BC² + AC²) - 1/2(AB² + BC² + AC²) = 2(AD² + FC² + BE²)
⇒3/2(AB² + BC² + AC²) = 2(AD² + FC² + BE²)
⇒3(AB² + BC² + AC²) = 4(AD² + FC² + BE²)
I HOPE ITS HELP YOU DEAR,
THANKS
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