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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 $AB$.

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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 $DB$.

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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 $I$.

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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.

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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.

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

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

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

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

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 $ABC$.

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  • 2
    $\begingroup$ Thanks! Very neat! I did not know the PoP theorem. Now I checked it. Sorry for the naivety, then! $\endgroup$ – user559615 Jul 30 '18 at 20:29
  • $\begingroup$ You don't need to be sorry? $\endgroup$ – Aqua Jul 30 '18 at 20:33

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