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Triangle-ABC-has-midpoints-D-E-and-F-By-connecting-each-verticie-with-the-opposite-midpoint-we-create-a-cress-section-called-G-Prove-that-all-three-lines-cross-at-point-G-regardless-of-the-type

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Question Number 4543 by FilupSmith last updated on 06/Feb/16
Triangle ABC has midpoints D, E and F. By connecting each verticie with the opposite midpoint, we create a cress−section called G. Prove that all three lines cross at point G regardless of the type of triangle
$$\mathrm{Triangle}\:{ABC}\:\mathrm{has}\:\mathrm{midpoints} \\ $$$${D},\:{E}\:\mathrm{and}\:{F}. \\ $$$$ \\ $$$$\mathrm{By}\:\mathrm{connecting}\:\mathrm{each}\:\mathrm{verticie}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{opposite}\:\mathrm{midpoint},\:\mathrm{we}\:\mathrm{create}\:\mathrm{a}\:\mathrm{cress}−\mathrm{section} \\ $$$$\mathrm{called}\:{G}. \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{all}\:\mathrm{three}\:\mathrm{lines}\:\mathrm{cross}\:\mathrm{at}\: \\ $$$$\mathrm{point}\:{G}\:\mathrm{regardless}\:\mathrm{of}\:\mathrm{the}\:\mathrm{type}\:\mathrm{of} \\ $$$$\mathrm{triangle} \\ $$
Commented by FilupSmith last updated on 06/Feb/16
Answered by Rasheed Soomro last updated on 06/Feb/16
Let A=(x_1 ,y_1 ) , B=(x_2 ,y_2 ) and C=(x_3 ,y_3 ) are vertices of the triangle. ∴ Midpoint of AB M_1 =( ((x_1 +x_2 )/2),((y_1 +y_2 )/2)) ∴ Midpoint of BC M_2 =( ((x_2 +x_3 )/2),((y_2 +y_3 )/2)) ∴ Midpoint of AC M_3 =( ((x_1 +x_3 )/2),((y_1 +y_3 )/2)) Now determine points G_1 ,G_2 ,G_3 which divide CM_1 AM_2 and BM_3 in 2:1 It will appear that G_1 =G_2 =G_3 =G=(((x_1 +x_2 +x_3 )/3),((y_1 +y_2 +y_3 )/3)) This proves not only that the medians are concurrent but also that they cut each other at ratio 2:1
$${Let}\:\mathrm{A}=\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{y}_{\mathrm{1}} \right)\:,\:\mathrm{B}=\left(\mathrm{x}_{\mathrm{2}} ,\mathrm{y}_{\mathrm{2}} \right)\:{and}\:\mathrm{C}=\left(\mathrm{x}_{\mathrm{3}} ,\mathrm{y}_{\mathrm{3}} \right) \\ $$$${are}\:{vertices}\:{of}\:{the}\:{triangle}. \\ $$$$\therefore\:{Midpoint}\:{of}\:\mathrm{AB}\:\:\:\mathrm{M}_{\mathrm{1}} =\left(\:\frac{\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} }{\mathrm{2}},\frac{\mathrm{y}_{\mathrm{1}} +\mathrm{y}_{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\therefore\:{Midpoint}\:{of}\:\mathrm{BC}\:\:\:\mathrm{M}_{\mathrm{2}} =\left(\:\frac{\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} }{\mathrm{2}},\frac{\mathrm{y}_{\mathrm{2}} +\mathrm{y}_{\mathrm{3}} }{\mathrm{2}}\right) \\ $$$$\therefore\:{Midpoint}\:{of}\:\mathrm{AC}\:\:\:\mathrm{M}_{\mathrm{3}} =\left(\:\frac{\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{3}} }{\mathrm{2}},\frac{\mathrm{y}_{\mathrm{1}} +\mathrm{y}_{\mathrm{3}} }{\mathrm{2}}\right) \\ $$$${Now}\:{determine}\:{points}\:\mathrm{G}_{\mathrm{1}} ,\mathrm{G}_{\mathrm{2}} ,\mathrm{G}_{\mathrm{3}} \:{which}\:{divide}\:\mathrm{CM}_{\mathrm{1}} \\ $$$$\mathrm{AM}_{\mathrm{2}} \:\:{and}\:\:\mathrm{BM}_{\mathrm{3}} \:{in}\:\:\mathrm{2}:\mathrm{1} \\ $$$${It}\:{will}\:{appear}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{G}_{\mathrm{1}} =\mathrm{G}_{\mathrm{2}} =\mathrm{G}_{\mathrm{3}} =\boldsymbol{\mathrm{G}}=\left(\frac{\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} }{\mathrm{3}},\frac{\mathrm{y}_{\mathrm{1}} +\mathrm{y}_{\mathrm{2}} +\mathrm{y}_{\mathrm{3}} }{\mathrm{3}}\right) \\ $$$${This}\:{proves}\:{not}\:{only}\:{that}\:{the}\:{medians}\:{are} \\ $$$${concurrent}\:{but}\:{also}\:{that}\:{they}\:{cut}\:{each}\:{other} \\ $$$${at}\:{ratio}\:\mathrm{2}:\mathrm{1} \\ $$$$ \\ $$

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