If the median from one vertex of a triangle is equal to the altitude from another vertex, then must the...












1












$begingroup$


Triangle$|BE|=|AD|$
and $|BD|=|DC|$



I know that in an equilateral triangle all auxiliary elements are equal. That is,



$$n_a=n_b=n_c=h_a=h_b=h_c=m_a=m_b=m_c$$



$m$, $n$, $h$ being median, angle bisector, and altitude respectively.




If $m_a=h_b$ in a triangle, does it imply that the triangle is equilateral (see picture)?











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  • $begingroup$
    Consider right triangle $triangle ABC$ with right angle at $C$, such that $|BC|=2p$ and $|AC|=psqrt{3}$. Note that $overline{BC}$ is the altitude from $B$, and that it is congruent to the median from $A$, yet the triangle is not equilateral (or even isosceles).
    $endgroup$
    – Blue
    Dec 5 '18 at 16:55


















1












$begingroup$


Triangle$|BE|=|AD|$
and $|BD|=|DC|$



I know that in an equilateral triangle all auxiliary elements are equal. That is,



$$n_a=n_b=n_c=h_a=h_b=h_c=m_a=m_b=m_c$$



$m$, $n$, $h$ being median, angle bisector, and altitude respectively.




If $m_a=h_b$ in a triangle, does it imply that the triangle is equilateral (see picture)?











share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider right triangle $triangle ABC$ with right angle at $C$, such that $|BC|=2p$ and $|AC|=psqrt{3}$. Note that $overline{BC}$ is the altitude from $B$, and that it is congruent to the median from $A$, yet the triangle is not equilateral (or even isosceles).
    $endgroup$
    – Blue
    Dec 5 '18 at 16:55
















1












1








1





$begingroup$


Triangle$|BE|=|AD|$
and $|BD|=|DC|$



I know that in an equilateral triangle all auxiliary elements are equal. That is,



$$n_a=n_b=n_c=h_a=h_b=h_c=m_a=m_b=m_c$$



$m$, $n$, $h$ being median, angle bisector, and altitude respectively.




If $m_a=h_b$ in a triangle, does it imply that the triangle is equilateral (see picture)?











share|cite|improve this question











$endgroup$




Triangle$|BE|=|AD|$
and $|BD|=|DC|$



I know that in an equilateral triangle all auxiliary elements are equal. That is,



$$n_a=n_b=n_c=h_a=h_b=h_c=m_a=m_b=m_c$$



$m$, $n$, $h$ being median, angle bisector, and altitude respectively.




If $m_a=h_b$ in a triangle, does it imply that the triangle is equilateral (see picture)?








geometry






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edited Dec 5 '18 at 16:38









Blue

47.9k870153




47.9k870153










asked Dec 5 '18 at 15:37









Eldar RahimliEldar Rahimli

1278




1278












  • $begingroup$
    Consider right triangle $triangle ABC$ with right angle at $C$, such that $|BC|=2p$ and $|AC|=psqrt{3}$. Note that $overline{BC}$ is the altitude from $B$, and that it is congruent to the median from $A$, yet the triangle is not equilateral (or even isosceles).
    $endgroup$
    – Blue
    Dec 5 '18 at 16:55




















  • $begingroup$
    Consider right triangle $triangle ABC$ with right angle at $C$, such that $|BC|=2p$ and $|AC|=psqrt{3}$. Note that $overline{BC}$ is the altitude from $B$, and that it is congruent to the median from $A$, yet the triangle is not equilateral (or even isosceles).
    $endgroup$
    – Blue
    Dec 5 '18 at 16:55


















$begingroup$
Consider right triangle $triangle ABC$ with right angle at $C$, such that $|BC|=2p$ and $|AC|=psqrt{3}$. Note that $overline{BC}$ is the altitude from $B$, and that it is congruent to the median from $A$, yet the triangle is not equilateral (or even isosceles).
$endgroup$
– Blue
Dec 5 '18 at 16:55






$begingroup$
Consider right triangle $triangle ABC$ with right angle at $C$, such that $|BC|=2p$ and $|AC|=psqrt{3}$. Note that $overline{BC}$ is the altitude from $B$, and that it is congruent to the median from $A$, yet the triangle is not equilateral (or even isosceles).
$endgroup$
– Blue
Dec 5 '18 at 16:55












1 Answer
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As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:



enter image description here



The construction is pretty straightforward: we draw a circle $Gamma_1$ centered at the midpoint of $BC$ with radius $ell$, a circle $Gamma_2$ centered at $B$ with radius $ell$, a tangent $tau$ from $C$ to $Gamma_2$. $A=Gamma_1captau$ solves the problem.



The red curve is an arc of the circle which goes through $M,C$ and the point $D$ on the perpendicular to $BC$ through $C$, such that $DM=BC$.



In particular $m_a=h_b$ implies that $AV=MC$, where $V$ is the "missing vertex" of an equilateral triangle built on $MC$, such that $V$ and $A$ lie on the same half-plane with respect to $BC$. Why? Because if $N$ is the projection of $M$ on $AC$, then $MN=frac{1}{2}BT=frac{1}{2}AM$ implies $widehat{MAN}=widehat{MAC}=30^{circ}$, hence $A$ lies on a circle centered at $V$ through $M$ and $C$.






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    $begingroup$

    As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:



    enter image description here



    The construction is pretty straightforward: we draw a circle $Gamma_1$ centered at the midpoint of $BC$ with radius $ell$, a circle $Gamma_2$ centered at $B$ with radius $ell$, a tangent $tau$ from $C$ to $Gamma_2$. $A=Gamma_1captau$ solves the problem.



    The red curve is an arc of the circle which goes through $M,C$ and the point $D$ on the perpendicular to $BC$ through $C$, such that $DM=BC$.



    In particular $m_a=h_b$ implies that $AV=MC$, where $V$ is the "missing vertex" of an equilateral triangle built on $MC$, such that $V$ and $A$ lie on the same half-plane with respect to $BC$. Why? Because if $N$ is the projection of $M$ on $AC$, then $MN=frac{1}{2}BT=frac{1}{2}AM$ implies $widehat{MAN}=widehat{MAC}=30^{circ}$, hence $A$ lies on a circle centered at $V$ through $M$ and $C$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:



      enter image description here



      The construction is pretty straightforward: we draw a circle $Gamma_1$ centered at the midpoint of $BC$ with radius $ell$, a circle $Gamma_2$ centered at $B$ with radius $ell$, a tangent $tau$ from $C$ to $Gamma_2$. $A=Gamma_1captau$ solves the problem.



      The red curve is an arc of the circle which goes through $M,C$ and the point $D$ on the perpendicular to $BC$ through $C$, such that $DM=BC$.



      In particular $m_a=h_b$ implies that $AV=MC$, where $V$ is the "missing vertex" of an equilateral triangle built on $MC$, such that $V$ and $A$ lie on the same half-plane with respect to $BC$. Why? Because if $N$ is the projection of $M$ on $AC$, then $MN=frac{1}{2}BT=frac{1}{2}AM$ implies $widehat{MAN}=widehat{MAC}=30^{circ}$, hence $A$ lies on a circle centered at $V$ through $M$ and $C$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:



        enter image description here



        The construction is pretty straightforward: we draw a circle $Gamma_1$ centered at the midpoint of $BC$ with radius $ell$, a circle $Gamma_2$ centered at $B$ with radius $ell$, a tangent $tau$ from $C$ to $Gamma_2$. $A=Gamma_1captau$ solves the problem.



        The red curve is an arc of the circle which goes through $M,C$ and the point $D$ on the perpendicular to $BC$ through $C$, such that $DM=BC$.



        In particular $m_a=h_b$ implies that $AV=MC$, where $V$ is the "missing vertex" of an equilateral triangle built on $MC$, such that $V$ and $A$ lie on the same half-plane with respect to $BC$. Why? Because if $N$ is the projection of $M$ on $AC$, then $MN=frac{1}{2}BT=frac{1}{2}AM$ implies $widehat{MAN}=widehat{MAC}=30^{circ}$, hence $A$ lies on a circle centered at $V$ through $M$ and $C$.






        share|cite|improve this answer











        $endgroup$



        As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:



        enter image description here



        The construction is pretty straightforward: we draw a circle $Gamma_1$ centered at the midpoint of $BC$ with radius $ell$, a circle $Gamma_2$ centered at $B$ with radius $ell$, a tangent $tau$ from $C$ to $Gamma_2$. $A=Gamma_1captau$ solves the problem.



        The red curve is an arc of the circle which goes through $M,C$ and the point $D$ on the perpendicular to $BC$ through $C$, such that $DM=BC$.



        In particular $m_a=h_b$ implies that $AV=MC$, where $V$ is the "missing vertex" of an equilateral triangle built on $MC$, such that $V$ and $A$ lie on the same half-plane with respect to $BC$. Why? Because if $N$ is the projection of $M$ on $AC$, then $MN=frac{1}{2}BT=frac{1}{2}AM$ implies $widehat{MAN}=widehat{MAC}=30^{circ}$, hence $A$ lies on a circle centered at $V$ through $M$ and $C$.







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        edited Dec 6 '18 at 1:03

























        answered Dec 5 '18 at 17:11









        Jack D'AurizioJack D'Aurizio

        289k33280660




        289k33280660






























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