Question

Q no 2 in the picture

please answer fast

Answer 1

Hy friend here is your answer

Given : AD, BE , CF be medians of ∆ ABC.

To Prove : 2( AD+ BE + CF)<3(AB+BC+CA)<4(AD+BE+CF)

Proof :

✍️In ∆ ABD ,

AD< AB+BD....1 ( In a triangle sum of two sides are greater then third one )

Similarly ,
In ∆ BCE

BE< BC + CE...(2){>>do<<}

In ∆ACF

CF < CA + AF.... {>>do<<}

Add (1),(2)&(3)...

AD+BE+CF<AB+BD+BC+CE+CA+AF

BD=1/2 BC
CE=1/2 CA
AF=1/2 AB
D,E,F are midpoint

Put the value

AD+BE+CF<AB+1/2AB+BC+1/2 BC+CA+1/2 CA

By LCM

2(AD+BE+CF)<2AB+AB+2BC+BC+2CA+CA

2(AD+BE+CF)<3AB+3BC+3CA

2(AD+BE+CF)<3(AB+BC+CA) ...(I)

✍️Let,

As G is the centroid, hence G divides AD, BE, CF in the ratio 2:1.

in the ratio 2:1.

BG = 2/3 BE,

CG = 2/3 CF,

Again in △BGC, 

BG + CG > BC

2/3BE + 2/3CF > BC
2BE + 2CF > 3BC Or 3BC < 2BE + 2CF ...........(i) 


Similarly, 
3CA < 2CF + 2AD...........(ii) 
3AB < 2AD + 2BE ..........(iii)

Add (i), (ii) and (iii) we get

3BC + 3CA + 3AB < 2BE + 2CF + 2CF + 2AD + 2AD +2BE

3(AB + BC + CA) < 4(AD + BE + CF)...(II)

From eqn. (I) & (II)

2(AD+BE+CF)<3(AB+BC+CA)< 4(AD + BE + CF)✅

☺️ I hope it's help you

Answer 2

Answer:

HOPE IT HELPS YOU FRIENDS