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


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Question Number 4543 by FilupSmith last updated on 06/Feb/16

$Triangle A B C has midpoints$ $D, E and F .$ $By connecting each verticie with the$ $opposite midpoint, we create a cress - section$ $called G .$ $Prove that all three lines cross at$ $point G regardless of the type of$ $triangle$
Commented by FilupSmith last updated on 06/Feb/16

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

	$L e t A = (x_{1}, y_{1}), B = (x_{2}, y_{2}) a n d C = (x_{3}, y_{3})$ $a r e v e r t i c e s o f t h e t r i a n g l e .$ $∴ M i d p o i n t o f AB M_{1} = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ $∴ M i d p o i n t o f BC M_{2} = (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2})$ $∴ M i d p o i n t o f AC M_{3} = (\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2})$ $N o w d e t e r m i n e p o i n t s G_{1}, G_{2}, G_{3} w h i c h d i v i d e {CM}_{1}$ ${AM}_{2} a n d {BM}_{3} i n 2 : 1$ $I t w i l l a p p e a r t h a t$ $G_{1} = G_{2} = G_{3} = G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ $T h i s p r o v e s n o t o n l y t h a t t h e m e d i a n s a r e$ $c o n c u r r e n t b u t a l s o t h a t t h e y c u t e a c h o t h e r$ $a t r a t i o 2 : 1$
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