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

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

Using vector method, prove that three

median of a triangle are concurrent .

Answered by MJS_new last updated on 12/May/21

let A = (\begin{matrix} 0 \\ 0 \end{matrix}) \land B = (\begin{matrix} c \\ 0 \end{matrix}) \land C = (\begin{matrix} p \\ h \end{matrix})

M_{a} = \frac{B + C}{2} = (\begin{matrix} \frac{c + p}{2} \\ \frac{h}{2} \end{matrix}) M_{b} = \frac{A + C}{2} = (\begin{matrix} \frac{p}{2} \\ \frac{h}{2} \end{matrix}) M_{c} = \frac{A + B}{2} = (\begin{matrix} \frac{c}{2} \\ 0 \end{matrix})

v_{a} = C - B = (\begin{matrix} p - c \\ h \end{matrix}) v_{b} = A - C = (\begin{matrix} - p \\ - h \end{matrix}) v_{c} = B - A = (\begin{matrix} c \\ 0 \end{matrix})

n_{a} = (\begin{matrix} - h \\ p - c \end{matrix}) n_{b} = (\begin{matrix} h \\ - p \end{matrix}) n_{c} = (\begin{matrix} 0 \\ c \end{matrix})

medians

m_{a} : X = M_{a} + λ_{a} n_{a} \Leftrightarrow y = \frac{c - p}{h} x - \frac{c^{2} - h^{2} - p^{2}}{2 h} (I)

m_{b} : X = M_{b} + λ_{b} n_{b} \Leftrightarrow y = - \frac{p}{h} x + \frac{h^{2} + p^{2}}{2 h} (I I)

m_{c} : X = M_{c} + λ_{c} n_{c} \Leftrightarrow x = \frac{c}{2}

inserting x = \frac{c}{2} in both equations I and I I we get

the same value for y

y = \frac{- c p + h^{2} + p^{2}}{2 h}

\Rightarrow m_{a} \cap m_{b} \cap m_{c} = (\begin{matrix} \frac{c}{2} \\ \frac{- c p + h^{2} + p^{2}}{2 h} \end{matrix})

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

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

T h e m e d i a n o f t r i a n g l e c u t s a n o t h e r

m e d i a n i n r a t i o (2 : 1)

f r o m A D \to E (\frac{\vec{B} + \vec{C}}{2})

P = \frac{1 \times A + \frac{2 (\vec{B} + \vec{C})}{2}}{3} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}

f r o m B E \to (\frac{\vec{A} + \vec{C}}{2})

P = \frac{1 \times B + \frac{2 (\vec{A} + \vec{C})}{2}}{3} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}

f r o m F C \to (\frac{\vec{A} + \vec{B}}{2})

P = \frac{1 \times C + \frac{2 (\vec{A} + \vec{B})}{2}}{3} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}

h e n c e i t s c o n c u r e n t