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Question Number 140735 by ZiYangLee last updated on 12/May/21

Using vector method, prove that three median of a triangle are concurrent.

$$\mathrm{Using}\:\mathrm{vector}\:\mathrm{method},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{three} \\ $$$$\mathrm{median}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{are}\:\mathrm{concurrent}. \\ $$

Answered by MJS_new last updated on 12/May/21

let A= ((0),(0) ) ∧B= ((c),(0) ) ∧C= ((p),(h) ) M_a =((B+C)/2)= ((((c+p)/2)),((h/2)) ) M_b =((A+C)/2)= (((p/2)),((h/2)) ) M_c =((A+B)/2)= (((c/2)),(0) ) v_a =C−B= (((p−c)),(h) ) v_b =A−C= (((−p)),((−h)) ) v_c =B−A= ((c),(0) ) n_a = (((−h)),((p−c)) ) n_b = ((h),((−p)) ) n_c = ((0),(c) ) medians m_a : X=M_a +λ_a n_a ⇔ y=((c−p)/h)x−((c^2 −h^2 −p^2 )/(2h)) (I) m_b : X=M_b +λ_b n_b ⇔ y=−(p/h)x+((h^2 +p^2 )/(2h)) (II) m_c : X=M_c +λ_c n_c ⇔ x=(c/2) inserting x=(c/2) in both equations I and II we get the same value for y y=((−cp+h^2 +p^2 )/(2h)) ⇒ m_a ∩m_b ∩m_c = (((c/2)),(((−cp+h^2 +p^2 )/(2h))) )

$$\mathrm{let}\:{A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\wedge{B}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix}\:\wedge{C}=\begin{pmatrix}{{p}}\\{{h}}\end{pmatrix} \\ $$$${M}_{{a}} =\frac{{B}+{C}}{\mathrm{2}}=\begin{pmatrix}{\frac{{c}+{p}}{\mathrm{2}}}\\{\frac{{h}}{\mathrm{2}}}\end{pmatrix}\:\:\:\:\:{M}_{{b}} =\frac{{A}+{C}}{\mathrm{2}}=\begin{pmatrix}{\frac{{p}}{\mathrm{2}}}\\{\frac{{h}}{\mathrm{2}}}\end{pmatrix}\:\:\:\:\:{M}_{{c}} =\frac{{A}+{B}}{\mathrm{2}}=\begin{pmatrix}{\frac{{c}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${v}_{{a}} ={C}−{B}=\begin{pmatrix}{{p}−{c}}\\{{h}}\end{pmatrix}\:\:\:\:\:{v}_{{b}} ={A}−{C}=\begin{pmatrix}{−{p}}\\{−{h}}\end{pmatrix}\:\:\:\:\:{v}_{{c}} ={B}−{A}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${n}_{{a}} =\begin{pmatrix}{−{h}}\\{{p}−{c}}\end{pmatrix}\:\:\:\:\:{n}_{{b}} =\begin{pmatrix}{{h}}\\{−{p}}\end{pmatrix}\:\:\:\:\:{n}_{{c}} =\begin{pmatrix}{\mathrm{0}}\\{{c}}\end{pmatrix} \\ $$$$\mathrm{medians} \\ $$$${m}_{{a}} :\:{X}={M}_{{a}} +\lambda_{{a}} {n}_{{a}} \:\Leftrightarrow\:{y}=\frac{{c}−{p}}{{h}}{x}−\frac{{c}^{\mathrm{2}} −{h}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{h}}\:\:\:\:\:\left({I}\right) \\ $$$${m}_{{b}} :\:{X}={M}_{{b}} +\lambda_{{b}} {n}_{{b}} \:\Leftrightarrow\:{y}=−\frac{{p}}{{h}}{x}+\frac{{h}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{h}}\:\:\:\:\:\left({II}\right) \\ $$$${m}_{{c}} :\:{X}={M}_{{c}} +\lambda_{{c}} {n}_{{c}} \:\Leftrightarrow\:{x}=\frac{{c}}{\mathrm{2}} \\ $$$$\mathrm{inserting}\:{x}=\frac{{c}}{\mathrm{2}}\:\mathrm{in}\:\mathrm{both}\:\mathrm{equations}\:{I}\:\mathrm{and}\:{II}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{value}\:\mathrm{for}\:{y} \\ $$$${y}=\frac{−{cp}+{h}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{h}} \\ $$$$\Rightarrow\:{m}_{{a}} \cap{m}_{{b}} \cap{m}_{{c}} =\begin{pmatrix}{\frac{{c}}{\mathrm{2}}}\\{\frac{−{cp}+{h}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{h}}}\end{pmatrix} \\ $$

Answered by peter frank last updated on 12/May/21

Commented by peter frank last updated on 12/May/21

The median of triangle cuts another median in ratio (2:1) from AD→E(((B^→ +C^→ )/2)) P=((1×A+((2(B^→ +C^→ ))/2))/3)=((A^→ +B^→ +C^→ )/3) from BE→(((A^→ +C^→ )/2)) P=((1×B+((2(A^→ +C^→ ))/2))/3)=((A^→ +B^→ +C^→ )/3) from FC→(((A^→ +B^→ )/2)) P=((1×C+((2(A^→ +B^→ ))/2))/3)=((A^→ +B^→ +C^→ )/3) hence its concurent

$${The}\:{median}\:{of}\:{triangle}\:{cuts}\:{another} \\ $$$${median}\:{in}\:{ratio}\:\left(\mathrm{2}:\mathrm{1}\right) \\ $$$${from}\:{AD}\rightarrow{E}\left(\frac{\overset{\rightarrow} {{B}}+\overset{\rightarrow} {{C}}}{\mathrm{2}}\right) \\ $$$${P}=\frac{\mathrm{1}×{A}+\frac{\mathrm{2}\left(\overset{\rightarrow} {{B}}+\overset{\rightarrow} {{C}}\right)}{\mathrm{2}}}{\mathrm{3}}=\frac{\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{B}}+\overset{\rightarrow} {{C}}}{\mathrm{3}} \\ $$$${from}\:{BE}\rightarrow\left(\frac{\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{C}}}{\mathrm{2}}\right) \\ $$$$\:\:{P}=\frac{\mathrm{1}×{B}+\frac{\mathrm{2}\left(\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{C}}\right)}{\mathrm{2}}}{\mathrm{3}}=\frac{\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{B}}+\overset{\rightarrow} {{C}}}{\mathrm{3}} \\ $$$${from}\:{FC}\rightarrow\left(\frac{\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{B}}}{\mathrm{2}}\right) \\ $$$${P}=\frac{\mathrm{1}×{C}+\frac{\mathrm{2}\left(\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{B}}\right)}{\mathrm{2}}}{\mathrm{3}}=\frac{\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{B}}+\overset{\rightarrow} {{C}}}{\mathrm{3}} \\ $$$${hence}\:{its}\:{concurent} \\ $$

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