Question

please solve
No Spams​

Answer 1

Answer:

Solution :-

Proof :-

(1) Let the three medians AD, BE and CF of ∆ABC intersect at the point G.

That is G is the centroid of ∆ABC.

∴ AG : GD = BG : GE = CG : GF = 2 : 1

Now, in ∆BGC, BG + CG > BC

in ∆AGC, AG + CG > CA

in ∆AGB, AG + BG > AB

So adding together, 2(AG + BG + CG) > AB + BC + CA

or, 2(2/3AD + 2/3BE + 2/3CF) > AB + BC + CA

or, 4/3(AD + BE + CF) > AB + BC + CA

or, 4(AD + BE + CF) > 3(AB + BC + CA)

∴ 4(AD + BE + CF) > 3(AB + BC + CA) (Proved).

(2) In ∆ABD, AB + BD > AD

In ∆BCE, BC + CE > BE

In ∆CAF, CA + AF > CF

So adding, (AD + BC + CA) + (BD + CE + AF) > (AD + BE + CF)

or, (AB + BC + CA) + (1/2BC + 1/2CA + 1/2AB) > AD + BE + CF

or, 3/2(AB + BC + CA) > AD + BE + CF

or, 3(AB + BC + CA) > 2(AD + BE + CF)

∴ 3(AB + BC + CA) > 2(AD + BE + CF) (Proved).

Answer 2

$\large\underline{\sf{Solution-1}}$

Given that AD, BE and CF be the medians of triangle ABC intersects BC, AC and AB at D, E and F respectively.

Let assume that medians AD, BE, CF intersects each other at G.

We know, Centroid of a triangle divides the medians in the ratio 2 : 1 from the vertex.

$\rm\implies \:AG : GD = BG : GE = CG : GF = 2 : 1$

It further implies

$\rm\implies \:AG = \frac{2}{3} AD$

$\rm\implies \:BG = \frac{2}{3}BE$

$\rm\implies \:CG = \frac{2}{3}CF$

Now, Consider

In triangle AGB

We know that sum of any two sides of a triangle is greater than third side.

$\rm \: AG + GB > AB$

On substituting the values of AG and BG from above, we get

$\rm \: \dfrac{2}{3}AD + \dfrac{2}{3}BE > AB$

$\rm\implies \:\rm \: 2AD + 2BE > 3AB - - - (1)$

Similarly,

$\rm\implies \:2AD + 2CF > 3AC - - - (2)$

Similarly,

$\rm\implies \:2BE + 2CF > 3BC - - - (3)$

On adding equation (1), (2) and (3), we get

$\rm\implies \:4AD + 4BE + 4CF > 3AC + 3AB + 3BC$

$\rm\implies \:4(AD + BE + CF) > 3(AB + BC + CA)$

$\color{green}\large\underline{\sf{Solution-2}}$

As AD is median. So, it bisects BC.

$\rm\implies \:BD = DC = \frac{1}{2} \: BC$

Now, In triangle ABD

We know, sum of any two sides of a triangle is greater than third side.

$\rm \: AB + BD > AD$

On substituting the value of BD from above, we get

$\rm \: AB + \frac{1}{2}BC > AD$

$\rm\implies \: \: 2AB + BC > 2AD - - - (1)$

Similarly,

$\rm\implies \:2BC + AC > 2BE - - - (2)$

Similarly,

$\rm\implies \:2AC + AB > 2CF - - - (3)$

So, on adding equation (1), (2) and (3), we get

$\rm \: 3AB + 3BC + 3AC > 2AD + 2BE + 2CF$

$\rm\implies \: 3(AB + BC + AC) > 2(AD + BE + CF)$

Hence, Proved