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Proving the orthocenter, circumcenter and centroid of a triangle are collinear.
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I just had a problem on a test related to Euler’s Line theorem,
(Euler Line Theorem: The orthocenter H, the circumcenter O, and the centroid G of any triangle are collinear. Furthermore, G is between H and O (unless the triangle is equilateral, in which case the three points coincide) and HG = 2GO.)
On the test problem, the professor gave us diagram of a triangle with three points in the interior of the triangle. The question said,
“Use the following diagram to prove synthetically that the circumcenter O, the centroid G, and the orthocenter H of a triangle are collinear.â€Â
Nobody in the class got full credit on it. He said it should be done this way:
Draw a line segment from C to G, and extend it such that CG=1/2 GH. Then prove that H is the orthocenter.
This didn’t sit well with me, and I don’t know what to think about his method.
1) Instead of a proof by contradiction, it seems like a proof by, I don’t know, acceptance? Accept that what you are trying to prove is true, then use half of what the theorem says to work backwards to prove something in the given. Is this a valid way to prove something? I have never heard of this, and it seems like it wouldn’t always be robust.
2) It rubs me the wrong way to be trying to prove that 3 points are collinear, and then your first step is to draw them so that they are collinear. It seems like a circular argument, although in the end he proved that H was the orthocenter. I like ending with what I am trying to prove, but maybe I am not right here.
3) Given the theorem, it seems like you are accepting that half of it is true, constructing them in this manner. If you are using half of the proof, why wouldn’t it be valid to accept the whole thing is true and say, “By the Euler Line theorem, these three points are collinear.â€Â
Any help would be greatly appreciated. This is really nagging at me.
euclidean-geometry
asked Jul 24 at 18:41
T.Jenkins
11
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up vote
0
down vote
favorite
I just had a problem on a test related to Euler’s Line theorem,
(Euler Line Theorem: The orthocenter H, the circumcenter O, and the centroid G of any triangle are collinear. Furthermore, G is between H and O (unless the triangle is equilateral, in which case the three points coincide) and HG = 2GO.)
On the test problem, the professor gave us diagram of a triangle with three points in the interior of the triangle. The question said,
“Use the following diagram to prove synthetically that the circumcenter O, the centroid G, and the orthocenter H of a triangle are collinear.â€Â
Nobody in the class got full credit on it. He said it should be done this way:
Draw a line segment from C to G, and extend it such that CG=1/2 GH. Then prove that H is the orthocenter.
This didn’t sit well with me, and I don’t know what to think about his method.
1) Instead of a proof by contradiction, it seems like a proof by, I don’t know, acceptance? Accept that what you are trying to prove is true, then use half of what the theorem says to work backwards to prove something in the given. Is this a valid way to prove something? I have never heard of this, and it seems like it wouldn’t always be robust.
2) It rubs me the wrong way to be trying to prove that 3 points are collinear, and then your first step is to draw them so that they are collinear. It seems like a circular argument, although in the end he proved that H was the orthocenter. I like ending with what I am trying to prove, but maybe I am not right here.
3) Given the theorem, it seems like you are accepting that half of it is true, constructing them in this manner. If you are using half of the proof, why wouldn’t it be valid to accept the whole thing is true and say, “By the Euler Line theorem, these three points are collinear.â€Â
Any help would be greatly appreciated. This is really nagging at me.
euclidean-geometry
asked Jul 24 at 18:41
T.Jenkins
11
1
Looks fine to me: you constructed a point $G$ and proved it was on all the altitudes. Thus $G$ is the orthocentre. Sounds good to me, and is constructive to boot.
– Lord Shark the Unknown
Jul 24 at 18:59
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I just had a problem on a test related to Euler’s Line theorem,
(Euler Line Theorem: The orthocenter H, the circumcenter O, and the centroid G of any triangle are collinear. Furthermore, G is between H and O (unless the triangle is equilateral, in which case the three points coincide) and HG = 2GO.)
On the test problem, the professor gave us diagram of a triangle with three points in the interior of the triangle. The question said,
“Use the following diagram to prove synthetically that the circumcenter O, the centroid G, and the orthocenter H of a triangle are collinear.â€Â
Nobody in the class got full credit on it. He said it should be done this way:
Draw a line segment from C to G, and extend it such that CG=1/2 GH. Then prove that H is the orthocenter.
This didn’t sit well with me, and I don’t know what to think about his method.
1) Instead of a proof by contradiction, it seems like a proof by, I don’t know, acceptance? Accept that what you are trying to prove is true, then use half of what the theorem says to work backwards to prove something in the given. Is this a valid way to prove something? I have never heard of this, and it seems like it wouldn’t always be robust.
2) It rubs me the wrong way to be trying to prove that 3 points are collinear, and then your first step is to draw them so that they are collinear. It seems like a circular argument, although in the end he proved that H was the orthocenter. I like ending with what I am trying to prove, but maybe I am not right here.
3) Given the theorem, it seems like you are accepting that half of it is true, constructing them in this manner. If you are using half of the proof, why wouldn’t it be valid to accept the whole thing is true and say, “By the Euler Line theorem, these three points are collinear.â€Â
Any help would be greatly appreciated. This is really nagging at me.
euclidean-geometry
asked Jul 24 at 18:41
T.Jenkins
11
I just had a problem on a test related to Euler’s Line theorem,
(Euler Line Theorem: The orthocenter H, the circumcenter O, and the centroid G of any triangle are collinear. Furthermore, G is between H and O (unless the triangle is equilateral, in which case the three points coincide) and HG = 2GO.)
On the test problem, the professor gave us diagram of a triangle with three points in the interior of the triangle. The question said,
“Use the following diagram to prove synthetically that the circumcenter O, the centroid G, and the orthocenter H of a triangle are collinear.â€Â
Nobody in the class got full credit on it. He said it should be done this way:
Draw a line segment from C to G, and extend it such that CG=1/2 GH. Then prove that H is the orthocenter.
This didn’t sit well with me, and I don’t know what to think about his method.
1) Instead of a proof by contradiction, it seems like a proof by, I don’t know, acceptance? Accept that what you are trying to prove is true, then use half of what the theorem says to work backwards to prove something in the given. Is this a valid way to prove something? I have never heard of this, and it seems like it wouldn’t always be robust.
2) It rubs me the wrong way to be trying to prove that 3 points are collinear, and then your first step is to draw them so that they are collinear. It seems like a circular argument, although in the end he proved that H was the orthocenter. I like ending with what I am trying to prove, but maybe I am not right here.
3) Given the theorem, it seems like you are accepting that half of it is true, constructing them in this manner. If you are using half of the proof, why wouldn’t it be valid to accept the whole thing is true and say, “By the Euler Line theorem, these three points are collinear.â€Â
Any help would be greatly appreciated. This is really nagging at me.
euclidean-geometry
asked Jul 24 at 18:41
T.Jenkins
11
asked Jul 24 at 18:41
T.Jenkins
11
asked Jul 24 at 18:41
T.Jenkins
11
asked Jul 24 at 18:41
T.Jenkins
11
11
1
Looks fine to me: you constructed a point $G$ and proved it was on all the altitudes. Thus $G$ is the orthocentre. Sounds good to me, and is constructive to boot.
– Lord Shark the Unknown
Jul 24 at 18:59
add a comment |Â
1
Looks fine to me: you constructed a point $G$ and proved it was on all the altitudes. Thus $G$ is the orthocentre. Sounds good to me, and is constructive to boot.
– Lord Shark the Unknown
Jul 24 at 18:59
1
1
Looks fine to me: you constructed a point $G$ and proved it was on all the altitudes. Thus $G$ is the orthocentre. Sounds good to me, and is constructive to boot.
– Lord Shark the Unknown
Jul 24 at 18:59
Looks fine to me: you constructed a point $G$ and proved it was on all the altitudes. Thus $G$ is the orthocentre. Sounds good to me, and is constructive to boot.
– Lord Shark the Unknown
Jul 24 at 18:59
add a comment |Â
2 Answers
2
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0
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Here is sketch of a different way to prove it.
Find the midlines of ABC.
The circumcenter of ABC is the othocenter of PQR
The centroid of ABC is the centroid of PQR
PQR is similar to ABC
Construct Euler line between the two orthocenter / Circumcenter of PQR / ABC and the Centroid, creating more similar triangles. And there are corresponding points between the othocenter of PQR and the orhtocenter of ABC along that line.
answered Jul 24 at 20:08
Doug M
39k31749
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up vote
0
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The proof you described is perfectly valid: it follows from the similitude of triangles $AGH$ and $MGO$ (see diagram, where $M$ is the midpoint of $BC$), which in turn implies $OMparallel AH$, so that $AH$ lies on the altitude from $A$ (and the same can reasoning can be repeated for $B$ and $C$). Triangles $AGH$ and $MGO$ are similar because $AG/MG=HG/OG=2$ and $angle AGH=angle OGM$.
Constructing point $H$ on line $OG$ and then showing that it is the orthocenter is indeed a very simple and clever way to avoid a direct collinearity proof. Compare it, for instance, with the proof given at cut-the-knot, which is more cumbersome.
answered Jul 25 at 17:45
Aretino
21.7k21342
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here is sketch of a different way to prove it.
Find the midlines of ABC.
The circumcenter of ABC is the othocenter of PQR
The centroid of ABC is the centroid of PQR
PQR is similar to ABC
Construct Euler line between the two orthocenter / Circumcenter of PQR / ABC and the Centroid, creating more similar triangles. And there are corresponding points between the othocenter of PQR and the orhtocenter of ABC along that line.
answered Jul 24 at 20:08
Doug M
39k31749
add a comment |Â
up vote
0
down vote
Here is sketch of a different way to prove it.
Find the midlines of ABC.
The circumcenter of ABC is the othocenter of PQR
The centroid of ABC is the centroid of PQR
PQR is similar to ABC
Construct Euler line between the two orthocenter / Circumcenter of PQR / ABC and the Centroid, creating more similar triangles. And there are corresponding points between the othocenter of PQR and the orhtocenter of ABC along that line.
answered Jul 24 at 20:08
Doug M
39k31749
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is sketch of a different way to prove it.
Find the midlines of ABC.
The circumcenter of ABC is the othocenter of PQR
The centroid of ABC is the centroid of PQR
PQR is similar to ABC
Construct Euler line between the two orthocenter / Circumcenter of PQR / ABC and the Centroid, creating more similar triangles. And there are corresponding points between the othocenter of PQR and the orhtocenter of ABC along that line.
answered Jul 24 at 20:08
Doug M
39k31749
Here is sketch of a different way to prove it.
Find the midlines of ABC.
The circumcenter of ABC is the othocenter of PQR
The centroid of ABC is the centroid of PQR
PQR is similar to ABC
Construct Euler line between the two orthocenter / Circumcenter of PQR / ABC and the Centroid, creating more similar triangles. And there are corresponding points between the othocenter of PQR and the orhtocenter of ABC along that line.
answered Jul 24 at 20:08
Doug M
39k31749
answered Jul 24 at 20:08
Doug M
39k31749
answered Jul 24 at 20:08
Doug M
39k31749
answered Jul 24 at 20:08
Doug M
39k31749
39k31749
add a comment |Â
add a comment |Â
up vote
0
down vote
The proof you described is perfectly valid: it follows from the similitude of triangles $AGH$ and $MGO$ (see diagram, where $M$ is the midpoint of $BC$), which in turn implies $OMparallel AH$, so that $AH$ lies on the altitude from $A$ (and the same can reasoning can be repeated for $B$ and $C$). Triangles $AGH$ and $MGO$ are similar because $AG/MG=HG/OG=2$ and $angle AGH=angle OGM$.
Constructing point $H$ on line $OG$ and then showing that it is the orthocenter is indeed a very simple and clever way to avoid a direct collinearity proof. Compare it, for instance, with the proof given at cut-the-knot, which is more cumbersome.
answered Jul 25 at 17:45
Aretino
21.7k21342
add a comment |Â
up vote
0
down vote
The proof you described is perfectly valid: it follows from the similitude of triangles $AGH$ and $MGO$ (see diagram, where $M$ is the midpoint of $BC$), which in turn implies $OMparallel AH$, so that $AH$ lies on the altitude from $A$ (and the same can reasoning can be repeated for $B$ and $C$). Triangles $AGH$ and $MGO$ are similar because $AG/MG=HG/OG=2$ and $angle AGH=angle OGM$.
Constructing point $H$ on line $OG$ and then showing that it is the orthocenter is indeed a very simple and clever way to avoid a direct collinearity proof. Compare it, for instance, with the proof given at cut-the-knot, which is more cumbersome.
answered Jul 25 at 17:45
Aretino
21.7k21342
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The proof you described is perfectly valid: it follows from the similitude of triangles $AGH$ and $MGO$ (see diagram, where $M$ is the midpoint of $BC$), which in turn implies $OMparallel AH$, so that $AH$ lies on the altitude from $A$ (and the same can reasoning can be repeated for $B$ and $C$). Triangles $AGH$ and $MGO$ are similar because $AG/MG=HG/OG=2$ and $angle AGH=angle OGM$.
Constructing point $H$ on line $OG$ and then showing that it is the orthocenter is indeed a very simple and clever way to avoid a direct collinearity proof. Compare it, for instance, with the proof given at cut-the-knot, which is more cumbersome.
answered Jul 25 at 17:45
Aretino
21.7k21342
The proof you described is perfectly valid: it follows from the similitude of triangles $AGH$ and $MGO$ (see diagram, where $M$ is the midpoint of $BC$), which in turn implies $OMparallel AH$, so that $AH$ lies on the altitude from $A$ (and the same can reasoning can be repeated for $B$ and $C$). Triangles $AGH$ and $MGO$ are similar because $AG/MG=HG/OG=2$ and $angle AGH=angle OGM$.
Constructing point $H$ on line $OG$ and then showing that it is the orthocenter is indeed a very simple and clever way to avoid a direct collinearity proof. Compare it, for instance, with the proof given at cut-the-knot, which is more cumbersome.
answered Jul 25 at 17:45
Aretino
21.7k21342
edited Jul 27 at 14:17
answered Jul 25 at 17:45
Aretino
21.7k21342
answered Jul 25 at 17:45
Aretino
21.7k21342
answered Jul 25 at 17:45
Aretino
21.7k21342
21.7k21342
add a comment |Â
add a comment |Â
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1
Looks fine to me: you constructed a point $G$ and proved it was on all the altitudes. Thus $G$ is the orthocentre. Sounds good to me, and is constructive to boot.
– Lord Shark the Unknown
Jul 24 at 18:59