Question

In ∆ABC, the median AD, BE and CF intersect at G. Show that 4(AD + BE + CF) > 3(AB + BC + AC)​

Answer 1

Step-by-step explanation:

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In a △ABC, medians AD,BE & CF intersect each other at G. Prove that.

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

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Solution

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

BG=

3

2

BEand

CG=

3

2

CF

Again in△BGC,

BG+CG>BC

⟹

3

2

BE+2CF>3BCor3BC<2BE+2CF−(1)

Similarly,3CA<2CF+2AD−(2)

3AB<2AD+2BE−(3)

By adding eqn(1),(2)&(3)

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

⟹3(AB+BC+CA)<4(AD+BE+CF)