Question

Prove that the sum of two sides of a triangle is greater than twice its median.

Spam and wrong answer will be deleted on the spot.

Answer 1

$\huge {\pink{\underline{\ulcorner{\star\: Answer\: \star }\urcorner}}}$

$\blue{\underline{ In \: \triangle ABC \: BO \: is\: a \:median}}$

$\bold{\red{\underline{Given\: \colon }}}$

∆ ABC and BO is a median and AO = OC

$\bold{\red{\underline{Construction\: \colon}}}$

By construction we extend C to B` so that AB || B`C extend BO to B` so , BO = OB` and AB = B`C or AB` = BC

$\bold{\red{\underline{Solution\: \colon}}}$

BB` = 2 BO ....{ BO = OB` }

............{ equation 01 }

$\red{\underline{In\: \triangle\:AB'B}}$

Statement 1 : The sum of two sides of triangle is always greater then third one.

So, in give triangle

AB + AB` > BB`

AB` = BC .........{by construction}

AB + BC > BB`

AB + BC > 2BO

{BO is a median and by equation 1 we put the value of BB`}

True statement : The sum of two triangle is always greater than the twice of the median of a triangle.

Hence ,

$\red{\large{\boxed{AB\: + BC \: > 2\:BO}}}$

Hence Proved

Answer 2

hy
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refer to the attachment !!

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hope helped
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