A conjecture about a circle bound to any triangle

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Given an equilateral triangle $ABC$, we choose a point $D$ inside it, determining a new triangle $ADB$.
enter image description here



We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $E$ and $F$ on the side $AB$.



enter image description here



If now we draw the two circles with center in $A$ and in $B$ and passing by $E$ and by $F$, respectively, we determine two new points $G$ and $H$ on the sides $AD$ and $DB$.



enter image description here



This post A conjecture related to a circle intrinsically bound to any triangle shows that the points $EGDHF$ determines always a circle.



Now we focus on the segments $DG$ and $CD$, and we draw their perpendicular bisectors. They intersect in the point $I$.



enter image description here



The circle with center in $I$ and passing by $C$, pass also through $G$ and $D$, for any $D$. Moreover, it always determines a point $J$ on the side $AC$ of the equilateral triangle.



enter image description here



A similar construction can be done starting from the perpendicular bisectors of $CD$ and $DH$, obtaining the center $K$ and the point $L$ on the side $CB$ of the equilateral triangle.



enter image description here




My conjecture is that the points $CJEFL$ always determine a circle.




enter image description here



Please, can you help me to find an elementary proof of such conjecture? Thanks for your suggestions!




geometry euclidean-geometry triangle




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    up vote
    8
    down vote

    favorite
    2












    Given an equilateral triangle $ABC$, we choose a point $D$ inside it, determining a new triangle $ADB$.
    enter image description here



    We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $E$ and $F$ on the side $AB$.



    enter image description here



    If now we draw the two circles with center in $A$ and in $B$ and passing by $E$ and by $F$, respectively, we determine two new points $G$ and $H$ on the sides $AD$ and $DB$.



    enter image description here



    This post A conjecture related to a circle intrinsically bound to any triangle shows that the points $EGDHF$ determines always a circle.



    Now we focus on the segments $DG$ and $CD$, and we draw their perpendicular bisectors. They intersect in the point $I$.



    enter image description here



    The circle with center in $I$ and passing by $C$, pass also through $G$ and $D$, for any $D$. Moreover, it always determines a point $J$ on the side $AC$ of the equilateral triangle.



    enter image description here



    A similar construction can be done starting from the perpendicular bisectors of $CD$ and $DH$, obtaining the center $K$ and the point $L$ on the side $CB$ of the equilateral triangle.



    enter image description here




    My conjecture is that the points $CJEFL$ always determine a circle.




    enter image description here



    Please, can you help me to find an elementary proof of such conjecture? Thanks for your suggestions!




    geometry euclidean-geometry triangle




    share|cite|improve this question





















      up vote
      8
      down vote

      favorite
      2









      up vote
      8
      down vote

      favorite
      2






      2





      Given an equilateral triangle $ABC$, we choose a point $D$ inside it, determining a new triangle $ADB$.
      enter image description here



      We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $E$ and $F$ on the side $AB$.



      enter image description here



      If now we draw the two circles with center in $A$ and in $B$ and passing by $E$ and by $F$, respectively, we determine two new points $G$ and $H$ on the sides $AD$ and $DB$.



      enter image description here



      This post A conjecture related to a circle intrinsically bound to any triangle shows that the points $EGDHF$ determines always a circle.



      Now we focus on the segments $DG$ and $CD$, and we draw their perpendicular bisectors. They intersect in the point $I$.



      enter image description here



      The circle with center in $I$ and passing by $C$, pass also through $G$ and $D$, for any $D$. Moreover, it always determines a point $J$ on the side $AC$ of the equilateral triangle.



      enter image description here



      A similar construction can be done starting from the perpendicular bisectors of $CD$ and $DH$, obtaining the center $K$ and the point $L$ on the side $CB$ of the equilateral triangle.



      enter image description here




      My conjecture is that the points $CJEFL$ always determine a circle.




      enter image description here



      Please, can you help me to find an elementary proof of such conjecture? Thanks for your suggestions!




      geometry euclidean-geometry triangle




      share|cite|improve this question











      Given an equilateral triangle $ABC$, we choose a point $D$ inside it, determining a new triangle $ADB$.
      enter image description here



      We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $E$ and $F$ on the side $AB$.



      enter image description here



      If now we draw the two circles with center in $A$ and in $B$ and passing by $E$ and by $F$, respectively, we determine two new points $G$ and $H$ on the sides $AD$ and $DB$.



      enter image description here



      This post A conjecture related to a circle intrinsically bound to any triangle shows that the points $EGDHF$ determines always a circle.



      Now we focus on the segments $DG$ and $CD$, and we draw their perpendicular bisectors. They intersect in the point $I$.



      enter image description here



      The circle with center in $I$ and passing by $C$, pass also through $G$ and $D$, for any $D$. Moreover, it always determines a point $J$ on the side $AC$ of the equilateral triangle.



      enter image description here



      A similar construction can be done starting from the perpendicular bisectors of $CD$ and $DH$, obtaining the center $K$ and the point $L$ on the side $CB$ of the equilateral triangle.



      enter image description here




      My conjecture is that the points $CJEFL$ always determine a circle.




      enter image description here



      Please, can you help me to find an elementary proof of such conjecture? Thanks for your suggestions!





      geometry euclidean-geometry triangle





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      asked Jul 30 at 19:40









      Andrea Prunotto

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          1 Answer
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          We see that $AE = AG$ and $AF = AD$.



          By PoP of $A$ with respect to circle $(CJGD)$ we have $$AJcdot AC = AG cdot AD = AE cdot AF$$



          so $J, C,E,F$ are concylic. The same is true for $L, C,E,F$ and we are done.



          Note: The statement is also true for arbitrary triangle $ABC$.






          share|cite|improve this answer



















          • 1




            Thanks! Very neat! I did not know the PoP theorem. Now I checked it. Sorry for the naivety, then!
            – Andrea Prunotto
            Jul 30 at 20:29










          • You don't need to be sorry?
            – greedoid
            Jul 30 at 20:33










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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          We see that $AE = AG$ and $AF = AD$.



          By PoP of $A$ with respect to circle $(CJGD)$ we have $$AJcdot AC = AG cdot AD = AE cdot AF$$



          so $J, C,E,F$ are concylic. The same is true for $L, C,E,F$ and we are done.



          Note: The statement is also true for arbitrary triangle $ABC$.






          share|cite|improve this answer



















          • 1




            Thanks! Very neat! I did not know the PoP theorem. Now I checked it. Sorry for the naivety, then!
            – Andrea Prunotto
            Jul 30 at 20:29










          • You don't need to be sorry?
            – greedoid
            Jul 30 at 20:33














          up vote
          2
          down vote



          accepted










          We see that $AE = AG$ and $AF = AD$.



          By PoP of $A$ with respect to circle $(CJGD)$ we have $$AJcdot AC = AG cdot AD = AE cdot AF$$



          so $J, C,E,F$ are concylic. The same is true for $L, C,E,F$ and we are done.



          Note: The statement is also true for arbitrary triangle $ABC$.






          share|cite|improve this answer



















          • 1




            Thanks! Very neat! I did not know the PoP theorem. Now I checked it. Sorry for the naivety, then!
            – Andrea Prunotto
            Jul 30 at 20:29










          • You don't need to be sorry?
            – greedoid
            Jul 30 at 20:33












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          We see that $AE = AG$ and $AF = AD$.



          By PoP of $A$ with respect to circle $(CJGD)$ we have $$AJcdot AC = AG cdot AD = AE cdot AF$$



          so $J, C,E,F$ are concylic. The same is true for $L, C,E,F$ and we are done.



          Note: The statement is also true for arbitrary triangle $ABC$.






          share|cite|improve this answer















          We see that $AE = AG$ and $AF = AD$.



          By PoP of $A$ with respect to circle $(CJGD)$ we have $$AJcdot AC = AG cdot AD = AE cdot AF$$



          so $J, C,E,F$ are concylic. The same is true for $L, C,E,F$ and we are done.



          Note: The statement is also true for arbitrary triangle $ABC$.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 30 at 20:12


























          answered Jul 30 at 20:02









          greedoid

          26k93473




          26k93473







          • 1




            Thanks! Very neat! I did not know the PoP theorem. Now I checked it. Sorry for the naivety, then!
            – Andrea Prunotto
            Jul 30 at 20:29










          • You don't need to be sorry?
            – greedoid
            Jul 30 at 20:33












          • 1




            Thanks! Very neat! I did not know the PoP theorem. Now I checked it. Sorry for the naivety, then!
            – Andrea Prunotto
            Jul 30 at 20:29










          • You don't need to be sorry?
            – greedoid
            Jul 30 at 20:33







          1




          1




          Thanks! Very neat! I did not know the PoP theorem. Now I checked it. Sorry for the naivety, then!
          – Andrea Prunotto
          Jul 30 at 20:29




          Thanks! Very neat! I did not know the PoP theorem. Now I checked it. Sorry for the naivety, then!
          – Andrea Prunotto
          Jul 30 at 20:29












          You don't need to be sorry?
          – greedoid
          Jul 30 at 20:33




          You don't need to be sorry?
          – greedoid
          Jul 30 at 20:33












           

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