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Question Number 207980 by mnjuly1970 last updated on 01/Jun/24

$$$$$$InA\stackrel{\mathrm{\Delta }}{BC}:B={90}^{\mathrm{o}}$$$${\mathrm{BB}}^{\prime }\mathrm{\perp }{\mathrm{CC}}^{\prime }\left({\mathrm{BB}}^{\prime }and{\mathrm{CC}}^{\prime }aremedians\right)$$$$$$$$\Rightarrow \frac{{\mathit{m}}_{\mathit{c}}}{{\mathit{m}}_{\mathit{a}}}=?$$$$\mathit{n}\mathit{o}\mathit{t}\mathit{e}:\mid {\mathrm{CC}}^{\prime }\mid ={\mathit{m}}_{\mathit{c}},\mid {\mathrm{AA}}^{\prime }\mid ={\mathit{m}}_{\mathit{a}}$$

Answered by mr W last updated on 02/Jun/24

Commented by mr W last updated on 02/Jun/24

$$B^{\prime }G=\frac{m_b}{3}=\frac{\sqrt{2(a^2+c^2)-b^2}}{6}$$$$CG=\frac{2m_c}{3}=\frac{\sqrt{2(a^2+b^2)-c^2}}{3}$$$$B^{\prime }C=\frac{b}{2}$$$${\left(\frac{\sqrt{2(a^2+c^2)-b^2}}{6}\right)}^2+{\left(\frac{\sqrt{2(a^2+b^2)-c^2}}{3}\right)}^2={\left(\frac{b}{2}\right)}^2$$$$\frac{2(a^2+c^2)-b^2}{36}+\frac{2(a^2+b^2)-c^2}{9}=\frac{b^2}{4}$$$$\Rightarrow 5a^2=b^2+c^2$$$$b^2=a^2+c^2$$$$\Rightarrow 2a^2=c^2$$$$\frac{m_c}{m_a}=\sqrt{\frac{2(a^2+b^2)-c^2}{2(b^2+c^2)-a^2}}$$$$=\sqrt{\frac{2(2a^2+2a^2)-2a^2}{2(a^2+4a^2)-a^2}}=\sqrt{\frac{2}{3}}✓$$

Commented by efronzo1 last updated on 02/Jun/24

$$\mathrm{AC}\mathrm{is}\cancel{\underbrace{}}$$

Commented by mr W last updated on 02/Jun/24

$$yes,since\mathrm{\angle }B=90°$$

Commented by mr W last updated on 02/Jun/24

Commented by mnjuly1970 last updated on 02/Jun/24

$$gratefulmaster$$$$veryniceasalways.\underset{―}{\underbrace{⋚}}$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{Weldone}\mathrm{sir}$$

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