Question

AD, BE and CF are medians of the triangle ABC, then prove that
3(AB² + BC² + CA²) = 4(AD² + BE² + CF²)

Answer 1

Answer:

Step-by-step explanation:

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

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 + BK²NOW,

⇒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 + BN²AND,

⇒BC² = CN² + BN²now,

⇒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)