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

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

Commented by FilupSmith last updated on 06/Feb/16

Answered by Rasheed Soomro last updated on 06/Feb/16

$$Let\mathrm{A}=({\mathrm{x}}_1,{\mathrm{y}}_1),\mathrm{B}=({\mathrm{x}}_2,{\mathrm{y}}_2)and\mathrm{C}=({\mathrm{x}}_3,{\mathrm{y}}_3)$$$$areverticesofthetriangle.$$$$\therefore Midpointof\mathrm{AB}{\mathrm{M}}_1=(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2})$$$$\therefore Midpointof\mathrm{BC}{\mathrm{M}}_2=(\frac{{\mathrm{x}}_2+{\mathrm{x}}_3}{2},\frac{{\mathrm{y}}_2+{\mathrm{y}}_3}{2})$$$$\therefore Midpointof\mathrm{AC}{\mathrm{M}}_3=(\frac{{\mathrm{x}}_1+{\mathrm{x}}_3}{2},\frac{{\mathrm{y}}_1+{\mathrm{y}}_3}{2})$$$$Nowdeterminepoints{\mathrm{G}}_1,{\mathrm{G}}_2,{\mathrm{G}}_{3}whichdivide{\mathrm{CM}}_1$$$${\mathrm{AM}}_{2}and{\mathrm{BM}}_{3}in2:1$$$$Itwillappearthat$$$${\mathrm{G}}_1={\mathrm{G}}_2={\mathrm{G}}_3=\text{boldsymbol}\mathrm{G}=(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3})$$$$Thisprovesnotonlythatthemediansare$$$$concurrentbutalsothattheycuteachother$$$$atratio2:1$$$$$$

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