If the median from one vertex of a triangle is equal to the altitude from another vertex, then must the...
$begingroup$
$|BE|=|AD|$
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
edited Dec 5 '18 at 16:38
Blue
47.7k870151
asked Dec 5 '18 at 15:37
Eldar RahimliEldar Rahimli
1088
$endgroup$
add a comment |
$begingroup$
$|BE|=|AD|$
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
edited Dec 5 '18 at 16:38
Blue
47.7k870151
asked Dec 5 '18 at 15:37
Eldar RahimliEldar Rahimli
1088
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$|BE|=|AD|$
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
edited Dec 5 '18 at 16:38
Blue
47.7k870151
asked Dec 5 '18 at 15:37
Eldar RahimliEldar Rahimli
1088
$endgroup$
$|BE|=|AD|$
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
geometry
edited Dec 5 '18 at 16:38
Blue
47.7k870151
asked Dec 5 '18 at 15:37
Eldar RahimliEldar Rahimli
1088
edited Dec 5 '18 at 16:38
Blue
47.7k870151
asked Dec 5 '18 at 15:37
Eldar RahimliEldar Rahimli
1088
edited Dec 5 '18 at 16:38
Blue
47.7k870151
edited Dec 5 '18 at 16:38
Blue
47.7k870151
edited Dec 5 '18 at 16:38
Blue
47.7k870151
47.7k870151
asked Dec 5 '18 at 15:37
Eldar RahimliEldar Rahimli
1088
asked Dec 5 '18 at 15:37
Eldar RahimliEldar Rahimli
1088
asked Dec 5 '18 at 15:37
Eldar RahimliEldar Rahimli
1088
1088
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
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$begingroup$
As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:
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 $C$.
answered Dec 5 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
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1 Answer
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$begingroup$
As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:
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 $C$.
answered Dec 5 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:
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 $C$.
answered Dec 5 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:
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 $C$.
answered Dec 5 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
As soon as the $A$-vertex lies on the red curve we have $m_a=h_b$:
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 $C$.
answered Dec 5 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
edited Dec 6 '18 at 1:03
answered Dec 5 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 5 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 5 '18 at 17:11
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
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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