Centroids of Euclidean triangles [duplicate]












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  • Why is the Centroid of a Triangle Shared with Its Midpoint Triangle

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Suppose that M,N, and P are the midpoints of sides AB, AC, and BC of triangle ABC. Show that the centroid of triangle MNP is the same point as the centroid of triangle ABC.










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marked as duplicate by Toby Mak, Nosrati, Lord Shark the Unknown, jgon, Leucippus Nov 19 '18 at 7:21


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    -4















    This question already has an answer here:




    • Why is the Centroid of a Triangle Shared with Its Midpoint Triangle

      2 answers




    Suppose that M,N, and P are the midpoints of sides AB, AC, and BC of triangle ABC. Show that the centroid of triangle MNP is the same point as the centroid of triangle ABC.










    share|cite|improve this question













    marked as duplicate by Toby Mak, Nosrati, Lord Shark the Unknown, jgon, Leucippus Nov 19 '18 at 7:21


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















      -4












      -4








      -4








      This question already has an answer here:




      • Why is the Centroid of a Triangle Shared with Its Midpoint Triangle

        2 answers




      Suppose that M,N, and P are the midpoints of sides AB, AC, and BC of triangle ABC. Show that the centroid of triangle MNP is the same point as the centroid of triangle ABC.










      share|cite|improve this question














      This question already has an answer here:




      • Why is the Centroid of a Triangle Shared with Its Midpoint Triangle

        2 answers




      Suppose that M,N, and P are the midpoints of sides AB, AC, and BC of triangle ABC. Show that the centroid of triangle MNP is the same point as the centroid of triangle ABC.





      This question already has an answer here:




      • Why is the Centroid of a Triangle Shared with Its Midpoint Triangle

        2 answers








      geometry






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      asked Nov 19 '18 at 3:26









      Jastys

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      marked as duplicate by Toby Mak, Nosrati, Lord Shark the Unknown, jgon, Leucippus Nov 19 '18 at 7:21


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Toby Mak, Nosrati, Lord Shark the Unknown, jgon, Leucippus Nov 19 '18 at 7:21


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























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          1.
          Consider the homothety which maps $∆ABC$ to its medial triangle $∆PNM$.



          HINT:



          What's the homothety center in such a construction?




          1. Note that by the inversion of Thale's Theorem (not to be confused with another theorem with the same name), since
            $$frac{overline{AM}}{overline{AB}}=frac{overline{AN}}{overline{AC}}$$
            the segment $overline{MN}$ is parallel to $overline{BC}$.


          Let $K=overline{AP}cap{overline{MN}}$. Again, by Thale's theorem
          $$frac{overline{AM}}{overline{AB}}=frac{overline{MK}}{overline{BP}}$$
          $$frac{overline{AN}}{overline{AC}}=frac{overline{KN}}{overline{PC}}$$
          $$Rightarrowfrac{overline{MK}}{overline{BP}}=frac{overline{KN}}{overline{PC}}Rightarrowfrac{overline{MK}}{overline{KN}}=frac{overline{BP}}{overline{PC}}=1$$
          This implies that K is also the midpoint of $overline{MN}$. Since $K$ lies on $overline{AP}$, $overline{AP}$ is one median in the triangle $∆PNM$. You can analugously prove this statement for $overline{BN}$ and $overline{CM}$. The centorid of $∆ABC$ is therefore the centroid of the triangle $∆PNM$






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            1 Answer
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            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            1.
            Consider the homothety which maps $∆ABC$ to its medial triangle $∆PNM$.



            HINT:



            What's the homothety center in such a construction?




            1. Note that by the inversion of Thale's Theorem (not to be confused with another theorem with the same name), since
              $$frac{overline{AM}}{overline{AB}}=frac{overline{AN}}{overline{AC}}$$
              the segment $overline{MN}$ is parallel to $overline{BC}$.


            Let $K=overline{AP}cap{overline{MN}}$. Again, by Thale's theorem
            $$frac{overline{AM}}{overline{AB}}=frac{overline{MK}}{overline{BP}}$$
            $$frac{overline{AN}}{overline{AC}}=frac{overline{KN}}{overline{PC}}$$
            $$Rightarrowfrac{overline{MK}}{overline{BP}}=frac{overline{KN}}{overline{PC}}Rightarrowfrac{overline{MK}}{overline{KN}}=frac{overline{BP}}{overline{PC}}=1$$
            This implies that K is also the midpoint of $overline{MN}$. Since $K$ lies on $overline{AP}$, $overline{AP}$ is one median in the triangle $∆PNM$. You can analugously prove this statement for $overline{BN}$ and $overline{CM}$. The centorid of $∆ABC$ is therefore the centroid of the triangle $∆PNM$






            share|cite|improve this answer


























              0














              1.
              Consider the homothety which maps $∆ABC$ to its medial triangle $∆PNM$.



              HINT:



              What's the homothety center in such a construction?




              1. Note that by the inversion of Thale's Theorem (not to be confused with another theorem with the same name), since
                $$frac{overline{AM}}{overline{AB}}=frac{overline{AN}}{overline{AC}}$$
                the segment $overline{MN}$ is parallel to $overline{BC}$.


              Let $K=overline{AP}cap{overline{MN}}$. Again, by Thale's theorem
              $$frac{overline{AM}}{overline{AB}}=frac{overline{MK}}{overline{BP}}$$
              $$frac{overline{AN}}{overline{AC}}=frac{overline{KN}}{overline{PC}}$$
              $$Rightarrowfrac{overline{MK}}{overline{BP}}=frac{overline{KN}}{overline{PC}}Rightarrowfrac{overline{MK}}{overline{KN}}=frac{overline{BP}}{overline{PC}}=1$$
              This implies that K is also the midpoint of $overline{MN}$. Since $K$ lies on $overline{AP}$, $overline{AP}$ is one median in the triangle $∆PNM$. You can analugously prove this statement for $overline{BN}$ and $overline{CM}$. The centorid of $∆ABC$ is therefore the centroid of the triangle $∆PNM$






              share|cite|improve this answer
























                0












                0








                0






                1.
                Consider the homothety which maps $∆ABC$ to its medial triangle $∆PNM$.



                HINT:



                What's the homothety center in such a construction?




                1. Note that by the inversion of Thale's Theorem (not to be confused with another theorem with the same name), since
                  $$frac{overline{AM}}{overline{AB}}=frac{overline{AN}}{overline{AC}}$$
                  the segment $overline{MN}$ is parallel to $overline{BC}$.


                Let $K=overline{AP}cap{overline{MN}}$. Again, by Thale's theorem
                $$frac{overline{AM}}{overline{AB}}=frac{overline{MK}}{overline{BP}}$$
                $$frac{overline{AN}}{overline{AC}}=frac{overline{KN}}{overline{PC}}$$
                $$Rightarrowfrac{overline{MK}}{overline{BP}}=frac{overline{KN}}{overline{PC}}Rightarrowfrac{overline{MK}}{overline{KN}}=frac{overline{BP}}{overline{PC}}=1$$
                This implies that K is also the midpoint of $overline{MN}$. Since $K$ lies on $overline{AP}$, $overline{AP}$ is one median in the triangle $∆PNM$. You can analugously prove this statement for $overline{BN}$ and $overline{CM}$. The centorid of $∆ABC$ is therefore the centroid of the triangle $∆PNM$






                share|cite|improve this answer












                1.
                Consider the homothety which maps $∆ABC$ to its medial triangle $∆PNM$.



                HINT:



                What's the homothety center in such a construction?




                1. Note that by the inversion of Thale's Theorem (not to be confused with another theorem with the same name), since
                  $$frac{overline{AM}}{overline{AB}}=frac{overline{AN}}{overline{AC}}$$
                  the segment $overline{MN}$ is parallel to $overline{BC}$.


                Let $K=overline{AP}cap{overline{MN}}$. Again, by Thale's theorem
                $$frac{overline{AM}}{overline{AB}}=frac{overline{MK}}{overline{BP}}$$
                $$frac{overline{AN}}{overline{AC}}=frac{overline{KN}}{overline{PC}}$$
                $$Rightarrowfrac{overline{MK}}{overline{BP}}=frac{overline{KN}}{overline{PC}}Rightarrowfrac{overline{MK}}{overline{KN}}=frac{overline{BP}}{overline{PC}}=1$$
                This implies that K is also the midpoint of $overline{MN}$. Since $K$ lies on $overline{AP}$, $overline{AP}$ is one median in the triangle $∆PNM$. You can analugously prove this statement for $overline{BN}$ and $overline{CM}$. The centorid of $∆ABC$ is therefore the centroid of the triangle $∆PNM$







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                answered Nov 19 '18 at 6:12









                Dr. Mathva

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