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Give-ABC-is-acute-triangle-M-is-a-midpoint-of-BC-Prove-that-AB-AC-gt-2AM-




Question Number 199481 by tri26112004 last updated on 04/Nov/23
Give △ABC is acute triangle.  M  is a midpoint of BC  Prove that AB+AC>2AM
$${Give}\:\bigtriangleup{ABC}\:{is}\:{acute}\:{triangle}. \\ $$$${M}\:\:{is}\:{a}\:{midpoint}\:{of}\:{BC} \\ $$$${Prove}\:{that}\:{AB}+{AC}>\mathrm{2}{AM} \\ $$
Answered by mr W last updated on 04/Nov/23
Commented by mr W last updated on 04/Nov/23
AB+BA′>AA′  ⇒AB+AC>2AM
$${AB}+{BA}'>{AA}' \\ $$$$\Rightarrow{AB}+{AC}>\mathrm{2}{AM} \\ $$
Commented by tri26112004 last updated on 04/Nov/23
you have other solution¿
$${you}\:{have}\:{other}\:{solution}¿ \\ $$
Commented by mr W last updated on 04/Nov/23
this is the simplest solution.
$${this}\:{is}\:{the}\:{simplest}\:{solution}. \\ $$
Commented by mr W last updated on 04/Nov/23
an other solution:  BC+AC>AB ⇒BC>AB−AC  2AM=(√(2AB^2 +2AC^2 −BC^2 ))             <(√(2AB^2 +2AC^2 −(AB−AC)^2 ))             =(√((AB+AC)^2 ))             =AB+AC
$${an}\:{other}\:{solution}: \\ $$$${BC}+{AC}>{AB}\:\Rightarrow{BC}>{AB}−{AC} \\ $$$$\mathrm{2}{AM}=\sqrt{\mathrm{2}{AB}^{\mathrm{2}} +\mathrm{2}{AC}^{\mathrm{2}} −{BC}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:<\sqrt{\mathrm{2}{AB}^{\mathrm{2}} +\mathrm{2}{AC}^{\mathrm{2}} −\left({AB}−{AC}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\left({AB}+{AC}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={AB}+{AC} \\ $$
Commented by tri26112004 last updated on 05/Nov/23
Good solution
$${Good}\:{solution} \\ $$

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