Parallelogram, 7th grade

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So, we have $ABCD$ a parallelogram, and three points $M$, $N$, $P$ on $AB$, $BC$ and $CD$ such that the center of the parallelogram is the center of gravity of $Delta MNP$. Show that N is the midle of BC. I have literally no idea. Thanks!!



I will be short, sorry, my time is very limited. I have to show that
$Delta CONequivDelta BON$, which is very simple.




geometry




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  • Can you do it when $ABCD$ is a rectangle?
    – Christoph
    Aug 14 at 8:38










  • Yeah, it s very easy like that!
    – Numbers
    Aug 14 at 8:39






  • 1




    Edit your question to show the proof for the case of rectangles and then think about how to apply the same argument to parallelograms.
    – Christoph
    Aug 14 at 8:41















up vote
2
down vote

favorite












So, we have $ABCD$ a parallelogram, and three points $M$, $N$, $P$ on $AB$, $BC$ and $CD$ such that the center of the parallelogram is the center of gravity of $Delta MNP$. Show that N is the midle of BC. I have literally no idea. Thanks!!



I will be short, sorry, my time is very limited. I have to show that
$Delta CONequivDelta BON$, which is very simple.




geometry




share|cite|improve this question






















  • Can you do it when $ABCD$ is a rectangle?
    – Christoph
    Aug 14 at 8:38










  • Yeah, it s very easy like that!
    – Numbers
    Aug 14 at 8:39






  • 1




    Edit your question to show the proof for the case of rectangles and then think about how to apply the same argument to parallelograms.
    – Christoph
    Aug 14 at 8:41













up vote
2
down vote

favorite









up vote
2
down vote

favorite











So, we have $ABCD$ a parallelogram, and three points $M$, $N$, $P$ on $AB$, $BC$ and $CD$ such that the center of the parallelogram is the center of gravity of $Delta MNP$. Show that N is the midle of BC. I have literally no idea. Thanks!!



I will be short, sorry, my time is very limited. I have to show that
$Delta CONequivDelta BON$, which is very simple.




geometry




share|cite|improve this question














So, we have $ABCD$ a parallelogram, and three points $M$, $N$, $P$ on $AB$, $BC$ and $CD$ such that the center of the parallelogram is the center of gravity of $Delta MNP$. Show that N is the midle of BC. I have literally no idea. Thanks!!



I will be short, sorry, my time is very limited. I have to show that
$Delta CONequivDelta BON$, which is very simple.





geometry





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 14 at 9:56









Daniel Fischer♦

172k16155274




172k16155274










asked Aug 14 at 8:35









Numbers

523




523











  • Can you do it when $ABCD$ is a rectangle?
    – Christoph
    Aug 14 at 8:38










  • Yeah, it s very easy like that!
    – Numbers
    Aug 14 at 8:39






  • 1




    Edit your question to show the proof for the case of rectangles and then think about how to apply the same argument to parallelograms.
    – Christoph
    Aug 14 at 8:41

















  • Can you do it when $ABCD$ is a rectangle?
    – Christoph
    Aug 14 at 8:38










  • Yeah, it s very easy like that!
    – Numbers
    Aug 14 at 8:39






  • 1




    Edit your question to show the proof for the case of rectangles and then think about how to apply the same argument to parallelograms.
    – Christoph
    Aug 14 at 8:41
















Can you do it when $ABCD$ is a rectangle?
– Christoph
Aug 14 at 8:38




Can you do it when $ABCD$ is a rectangle?
– Christoph
Aug 14 at 8:38












Yeah, it s very easy like that!
– Numbers
Aug 14 at 8:39




Yeah, it s very easy like that!
– Numbers
Aug 14 at 8:39




1




1




Edit your question to show the proof for the case of rectangles and then think about how to apply the same argument to parallelograms.
– Christoph
Aug 14 at 8:41





Edit your question to show the proof for the case of rectangles and then think about how to apply the same argument to parallelograms.
– Christoph
Aug 14 at 8:41











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Extend $NT$ to $Ein MP$. Then $E$ halves $PM$, so $$d(E,AB) = 1over 2d(AB,CD)$$
and the sam is true for $T$ (since diagonals in parallelogram halves each other): $$d(T,AB) = 1over 2d(AB,CD)$$



So $ET||AB$ and thus $EM||AB$ so: $$d(N,AB) = 1over 2d(AB,CD)$$ and thus $N$ halves $BC$.






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    Extend $NT$ to $Ein MP$. Then $E$ halves $PM$, so $$d(E,AB) = 1over 2d(AB,CD)$$
    and the sam is true for $T$ (since diagonals in parallelogram halves each other): $$d(T,AB) = 1over 2d(AB,CD)$$



    So $ET||AB$ and thus $EM||AB$ so: $$d(N,AB) = 1over 2d(AB,CD)$$ and thus $N$ halves $BC$.






    share|cite|improve this answer
























      up vote
      1
      down vote













      Extend $NT$ to $Ein MP$. Then $E$ halves $PM$, so $$d(E,AB) = 1over 2d(AB,CD)$$
      and the sam is true for $T$ (since diagonals in parallelogram halves each other): $$d(T,AB) = 1over 2d(AB,CD)$$



      So $ET||AB$ and thus $EM||AB$ so: $$d(N,AB) = 1over 2d(AB,CD)$$ and thus $N$ halves $BC$.






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        Extend $NT$ to $Ein MP$. Then $E$ halves $PM$, so $$d(E,AB) = 1over 2d(AB,CD)$$
        and the sam is true for $T$ (since diagonals in parallelogram halves each other): $$d(T,AB) = 1over 2d(AB,CD)$$



        So $ET||AB$ and thus $EM||AB$ so: $$d(N,AB) = 1over 2d(AB,CD)$$ and thus $N$ halves $BC$.






        share|cite|improve this answer












        Extend $NT$ to $Ein MP$. Then $E$ halves $PM$, so $$d(E,AB) = 1over 2d(AB,CD)$$
        and the sam is true for $T$ (since diagonals in parallelogram halves each other): $$d(T,AB) = 1over 2d(AB,CD)$$



        So $ET||AB$ and thus $EM||AB$ so: $$d(N,AB) = 1over 2d(AB,CD)$$ and thus $N$ halves $BC$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 14 at 10:12









        greedoid

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