Question

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Here is my question in the pic.

Q.22
Need detailed answer .

Chapter - Pythagoras Theorem

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Answer 1

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)


$\bold{\boxed{HENCE, 3(AB^2 + BC^2 + AC^2) = 4(AD^2 + FC^2 + BE^2)}}$

$\mathit{\large{I\:\: HOPE\:\: IT'S\:\: HELP\:\: YOU\:\: DEAR,}}\\\mathit{\large{THANKS}}$

Answer 2

I skipped some minor Calculations, Hope you can easily Figure it out.

First we will Prove :
$AB^2 + AC^2 = 2 AD^2 + \frac{1}{2}BC^2$

Steps:
1) Let ABC be a triangle in which AD is median .
Construction : Draw $AE \perp BC.$

Proof : By Pythagoras Theorem, :
In right $\Delta ABE \\ AB^2 = AE^2 + BE^2$

In right $\Delta ADE, \\ AD^2 = AE^2 + ED^2$

In right, $\Delta AEC, \\ AC^2 = AE^2 + EC^2 \\ \\ 2) Now, \\ LHS = AB^2 + AC^2 \\ =[AE^2 + BE^2] + [AE^2 + EC^2 ] \\ = 2AE^2 + (BD-DE) ^2 + (CD+DE)^2 \\ = 2 AE^2 + 2(BD^2 + DE^2) \\ = 2 (AE^2 +DE^2) + 2 BD^2 \\ = 2 AD^2 + 2 (BC/2)^2 \: (as \:BD = BC/2)\\ AB^2+ BC^2 = 2 AD^2 + BC^2/2 --(1)$

Hence, We proved that sum of squares of any two sides of a triangle is equal to sum of square of median drawn to third side plus half the square of third side.

3) Now, Coming to the Question :D
We have other medians like BE & CF .

Similarly as above we can say,
$AB^2 + BC^2 = 2 BE^2 + AC^2/2 ---(2) \\ BC^2 + AC^2 = 2 CF^2 + AB^2/2 ---(3)$

4) Add all equations (1), (2) & (3),
$2( AB^2+ BC^2+ AC^2) = 2(AD^2+ BE^2+CF^2)+(AB^2+BC^2+AC^2)/2 \\ \\ => (2-1/2)[AB^2+BC^2+AC^2) = 2(AD^2+BE^2+CF^2)\\ => 3(AB^2+BC^2+AC^2) = 4(AD^2+BE^2+CF^2)$

Hence Proved :