S-4-3-m-m-m-a-m-m-b-m-m-c-m-m-a-m-b-m-c-2-m-a-m-b-m-c-mediani-prove-

Question Number 148821 by vvvv last updated on 31/Jul/21

$$\boldsymbol{{S}}=\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\boldsymbol{{m}}\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{a}}} \right)\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{b}}} \right)\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{c}}} \right)} \\ $$$$\boldsymbol{{m}}=\frac{\boldsymbol{{m}}_{\boldsymbol{{a}}} +\boldsymbol{{m}}_{\boldsymbol{{b}}} +\boldsymbol{{m}}_{\boldsymbol{{c}}} }{\mathrm{2}} \\ $$$$\boldsymbol{{m}}_{\boldsymbol{{a}}} ;\boldsymbol{{m}}_{\boldsymbol{{b}}} ;\boldsymbol{{m}}_{\boldsymbol{{c}}} −\boldsymbol{{mediani}} \\ $$$$\boldsymbol{{prove}} \\ $$

Answered by mr W last updated on 31/Jul/21

Commented by mr W last updated on 01/Aug/21

$${i}\:{don}'{t}\:{know}. \\ $$

Commented by mr W last updated on 31/Jul/21

proof in short: AG=(2/3)AD, BE=3×GE AD=m_a , DK=BE=m_b , AK=CF=m_c Δ_(ADK) =(√(m(m−m_a )(m−m_b )(m−m_c ))) Δ_(AGC) =Δ_(AGH) =((2/3))^2 Δ_(ADK) S=Δ_(ABC) =3Δ_(AGC) =(4/3)(√(m(m−m_a )(m−m_b )(m−m_c )))

$${proof}\:{in}\:{short}: \\ $$$${AG}=\frac{\mathrm{2}}{\mathrm{3}}{AD},\:{BE}=\mathrm{3}×{GE} \\ $$$${AD}={m}_{{a}} ,\:{DK}={BE}={m}_{{b}} ,\:{AK}={CF}={m}_{{c}} \\ $$$$\Delta_{{ADK}} =\sqrt{{m}\left({m}−{m}_{{a}} \right)\left({m}−{m}_{{b}} \right)\left({m}−{m}_{{c}} \right)} \\ $$$$\Delta_{{AGC}} =\Delta_{{AGH}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \Delta_{{ADK}} \\ $$$${S}=\Delta_{{ABC}} =\mathrm{3}\Delta_{{AGC}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\sqrt{{m}\left({m}−{m}_{{a}} \right)\left({m}−{m}_{{b}} \right)\left({m}−{m}_{{c}} \right)} \\ $$

Commented by vvvv last updated on 31/Jul/21