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

$$\mathit{S}=\frac{4}{3}\sqrt{\mathit{m}(\mathit{m}-{\mathit{m}}_{\mathit{a}})(\mathit{m}-{\mathit{m}}_{\mathit{b}})(\mathit{m}-{\mathit{m}}_{\mathit{c}})}$$$$\mathit{m}=\frac{{\mathit{m}}_{\mathit{a}}+{\mathit{m}}_{\mathit{b}}+{\mathit{m}}_{\mathit{c}}}{2}$$$${\mathit{m}}_{\mathit{a}};{\mathit{m}}_{\mathit{b}};{\mathit{m}}_{\mathit{c}}-\mathit{m}\mathit{e}\mathit{d}\mathit{i}\mathit{a}\mathit{n}\mathit{i}$$$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}$$

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

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

$$i{don}^{\prime }tknow.$$

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

$$proofinshort:$$$$AG=\frac{2}{3}AD,BE=3\times GE$$$$AD=m_a,DK=BE=m_b,AK=CF=m_c$$$${\mathrm{\Delta }}_{ADK}=\sqrt{m(m-m_a)(m-m_b)(m-m_c)}$$$${\mathrm{\Delta }}_{AGC}={\mathrm{\Delta }}_{AGH}={\left(\frac{2}{3}\right)}^2{\mathrm{\Delta }}_{ADK}$$$$S={\mathrm{\Delta }}_{ABC}=3{\mathrm{\Delta }}_{AGC}$$$$=\frac{4}{3}\sqrt{m(m-m_a)(m-m_b)(m-m_c)}$$

Commented by vvvv last updated on 31/Jul/21

$$\mathit{C}\mathit{K}\ne {\mathit{m}}_{\mathit{c}}\mathit{C}\mathit{F}={\mathit{m}}_{\mathit{c}}$$

Commented by vvvv last updated on 31/Jul/21

$$drawsoftwarenameplease$$

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

$$LEKHDIAGRAM$$

Commented by EDWIN88 last updated on 01/Aug/21

$$compatibletopc?$$

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