Mixing
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Proving $|CH| = 2|OM|$ of $triangle ABC$ of which $O$ is the circumcenter and $H$ is the orthocenter.
$begingroup$
Let $ABC$ be a triangle as shown in the figure below, where $O$ is its circumcenter and $H$ is its orthocenter. $AB$ is the opposite side of the climax point $C$, and $OM$ is perpendicular to $AB$.
How to prove that $|CH| = 2|OM|$?
$G$ is the centroid of the triangle. As the circumcenter, centroid and orthocenter (respectively $O$, $G$, and $H$) of the circle is collinear, we can get two specific triangle $CGH$ and $OMG$.
Showing both the triangle similar to each other by their respective angle, we can get the proportion of their 2 different length. If we consider the point $G$, then we can get $|CG|:|GM| = 2:1$. Thus we can easily prove this.
However, we can reverse the proof showing that $|CH|$ is equal to $2|OM|$ (It is included in our textbook as a corollary or something other). If we go with this process, it will lead to something contradictory.
Because of claiming the $G$ as centroid of the triangle, we have already used the above condition which needs to be proved.
So, is there any other method to prove that $|CH|=2|OM|$?
proof-verification euclidean-geometry triangle plane-geometry
asked Jan 22 at 11:12
Anirban NiloyAnirban Niloy
8291218
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be a triangle as shown in the figure below, where $O$ is its circumcenter and $H$ is its orthocenter. $AB$ is the opposite side of the climax point $C$, and $OM$ is perpendicular to $AB$.
How to prove that $|CH| = 2|OM|$?
$G$ is the centroid of the triangle. As the circumcenter, centroid and orthocenter (respectively $O$, $G$, and $H$) of the circle is collinear, we can get two specific triangle $CGH$ and $OMG$.
Showing both the triangle similar to each other by their respective angle, we can get the proportion of their 2 different length. If we consider the point $G$, then we can get $|CG|:|GM| = 2:1$. Thus we can easily prove this.
However, we can reverse the proof showing that $|CH|$ is equal to $2|OM|$ (It is included in our textbook as a corollary or something other). If we go with this process, it will lead to something contradictory.
Because of claiming the $G$ as centroid of the triangle, we have already used the above condition which needs to be proved.
So, is there any other method to prove that $|CH|=2|OM|$?
proof-verification euclidean-geometry triangle plane-geometry
asked Jan 22 at 11:12
Anirban NiloyAnirban Niloy
8291218
$endgroup$
$begingroup$
If $CH=2OM$ was proved while proving another equality, then you already have an independent proof of that.
$endgroup$
– Aretino
Jan 22 at 14:14
$begingroup$
I think, I have been futile to clearify my text. While proving another equality, I just used CH = 2OM as a preknowledge, not was proved in that proof. So I wanted a distinctive proof of above condition which has been proposed.
$endgroup$
– Anirban Niloy
Jan 22 at 14:22
add a comment |
$begingroup$
Let $ABC$ be a triangle as shown in the figure below, where $O$ is its circumcenter and $H$ is its orthocenter. $AB$ is the opposite side of the climax point $C$, and $OM$ is perpendicular to $AB$.
How to prove that $|CH| = 2|OM|$?
$G$ is the centroid of the triangle. As the circumcenter, centroid and orthocenter (respectively $O$, $G$, and $H$) of the circle is collinear, we can get two specific triangle $CGH$ and $OMG$.
Showing both the triangle similar to each other by their respective angle, we can get the proportion of their 2 different length. If we consider the point $G$, then we can get $|CG|:|GM| = 2:1$. Thus we can easily prove this.
However, we can reverse the proof showing that $|CH|$ is equal to $2|OM|$ (It is included in our textbook as a corollary or something other). If we go with this process, it will lead to something contradictory.
Because of claiming the $G$ as centroid of the triangle, we have already used the above condition which needs to be proved.
So, is there any other method to prove that $|CH|=2|OM|$?
proof-verification euclidean-geometry triangle plane-geometry
asked Jan 22 at 11:12
Anirban NiloyAnirban Niloy
8291218
$endgroup$
Let $ABC$ be a triangle as shown in the figure below, where $O$ is its circumcenter and $H$ is its orthocenter. $AB$ is the opposite side of the climax point $C$, and $OM$ is perpendicular to $AB$.
How to prove that $|CH| = 2|OM|$?
$G$ is the centroid of the triangle. As the circumcenter, centroid and orthocenter (respectively $O$, $G$, and $H$) of the circle is collinear, we can get two specific triangle $CGH$ and $OMG$.
Showing both the triangle similar to each other by their respective angle, we can get the proportion of their 2 different length. If we consider the point $G$, then we can get $|CG|:|GM| = 2:1$. Thus we can easily prove this.
However, we can reverse the proof showing that $|CH|$ is equal to $2|OM|$ (It is included in our textbook as a corollary or something other). If we go with this process, it will lead to something contradictory.
Because of claiming the $G$ as centroid of the triangle, we have already used the above condition which needs to be proved.
So, is there any other method to prove that $|CH|=2|OM|$?
proof-verification euclidean-geometry triangle plane-geometry
proof-verification euclidean-geometry triangle plane-geometry
asked Jan 22 at 11:12
Anirban NiloyAnirban Niloy
8291218
asked Jan 22 at 11:12
Anirban NiloyAnirban Niloy
8291218
edited Feb 11 at 17:40
Anirban Niloy
asked Jan 22 at 11:12
Anirban NiloyAnirban Niloy
8291218
asked Jan 22 at 11:12
Anirban NiloyAnirban Niloy
8291218
asked Jan 22 at 11:12
Anirban NiloyAnirban Niloy
8291218
8291218
$begingroup$
If $CH=2OM$ was proved while proving another equality, then you already have an independent proof of that.
$endgroup$
– Aretino
Jan 22 at 14:14
$begingroup$
I think, I have been futile to clearify my text. While proving another equality, I just used CH = 2OM as a preknowledge, not was proved in that proof. So I wanted a distinctive proof of above condition which has been proposed.
$endgroup$
– Anirban Niloy
Jan 22 at 14:22
add a comment |
$begingroup$
If $CH=2OM$ was proved while proving another equality, then you already have an independent proof of that.
$endgroup$
– Aretino
Jan 22 at 14:14
$begingroup$
I think, I have been futile to clearify my text. While proving another equality, I just used CH = 2OM as a preknowledge, not was proved in that proof. So I wanted a distinctive proof of above condition which has been proposed.
$endgroup$
– Anirban Niloy
Jan 22 at 14:22
$begingroup$
If $CH=2OM$ was proved while proving another equality, then you already have an independent proof of that.
$endgroup$
– Aretino
Jan 22 at 14:14
$begingroup$
If $CH=2OM$ was proved while proving another equality, then you already have an independent proof of that.
$endgroup$
– Aretino
Jan 22 at 14:14
$begingroup$
I think, I have been futile to clearify my text. While proving another equality, I just used CH = 2OM as a preknowledge, not was proved in that proof. So I wanted a distinctive proof of above condition which has been proposed.
$endgroup$
– Anirban Niloy
Jan 22 at 14:22
$begingroup$
I think, I have been futile to clearify my text. While proving another equality, I just used CH = 2OM as a preknowledge, not was proved in that proof. So I wanted a distinctive proof of above condition which has been proposed.
$endgroup$
– Anirban Niloy
Jan 22 at 14:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The most ellegant proof (at least in my opinion) is vector based:
Vectors $vec{CH}$ and $vec{OM}$ are parallel as well as vectors $vec{AH}$ and $vec{ON}$ and therefore:
$$vec{OM}=alphavec{CH}$$
$$vec{ON}=betavec{AH}$$
...for some real values of $alpha,beta$.
Segment MN connects midpoints of sides $AB$ and $BC$ and therefore:
$$vec{MN}=frac12vec{AC}=frac12(vec{AH}+vec{HC})=frac12vec{AH}-frac12vec{CH}tag{1}$$
On the other side it is obvious that:
$$vec{MN}=vec{MO}+vec{ON}=-vec{OM}+vec{ON}=-alphavec{CH}+betavec{AH}tag{2}$$
From (1) and (2) you get that:
$$frac12vec{AH}-frac12vec{CH}=-alphavec{CH}+betavec{AH}$$
$$(frac12-beta)vec{AH}+(alpha-frac12)vec{CH}=0tag{3}$$
Vectors $vec{AH}$ and $vec{CH}$ are not parallel! If you multiply these vectors with some real factors different from zero, you still get vectors that are not parallel (vectors are just stretched). Sum of two vectors that are not parallel cannot be zero. So (3) can be valid only if both scalars are equal to zero:
$$frac12 - beta =0, alpha-frac12=0$$
...which simply means that $alpha=beta=frac12$. Consequentially:
$$vec{OM}=frac12vec{CH}$$
$$OM=frac12 CH$$
answered Jan 22 at 13:50
OldboyOldboy
8,62711036
$endgroup$
$begingroup$
Outstanding solution, Oldboy. But my question is that which term is better for AH and CH vector? Collinearity or obliqueness? I didn't understand and I need more clarification about what you said that if we multiply (3) vectors with some real factors different from zero, we still get vectors that aren't collinear. If you get time, please explain that.
$endgroup$
– Anirban Niloy
Jan 22 at 14:43
1
$begingroup$
I think that you are right. I have used "collinear" while in fact I meant "parallel". I'm going to correct that. Basically, if you have two vectors that are not parallel their sum cannot be zero! Multiply these two vectors with some scalars, it won't change their orientation in space. Non-parallel vectors will still be non-parallel and their sum cannot be zero.
$endgroup$
– Oldboy
Jan 22 at 14:54
add a comment |
$begingroup$
Do you know Hamilton theorem? It says $$vec{OH}= vec{OA}+vec{OB}+vec{OC} $$
So $$vec{CH}=vec{OH}- vec{OC}= vec{OA}+vec{OB} ={1over 2}vec{OM} $$
answered Feb 4 at 16:58
Maria MazurMaria Mazur
46.4k1160119
$endgroup$
$begingroup$
No, I didn't know about it. You helped me to learn a new thing which is more appreciated. If possible, I will try to read about the personal details of Hamilton theorem.
$endgroup$
– Anirban Niloy
Feb 5 at 1:38
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The most ellegant proof (at least in my opinion) is vector based:
Vectors $vec{CH}$ and $vec{OM}$ are parallel as well as vectors $vec{AH}$ and $vec{ON}$ and therefore:
$$vec{OM}=alphavec{CH}$$
$$vec{ON}=betavec{AH}$$
...for some real values of $alpha,beta$.
Segment MN connects midpoints of sides $AB$ and $BC$ and therefore:
$$vec{MN}=frac12vec{AC}=frac12(vec{AH}+vec{HC})=frac12vec{AH}-frac12vec{CH}tag{1}$$
On the other side it is obvious that:
$$vec{MN}=vec{MO}+vec{ON}=-vec{OM}+vec{ON}=-alphavec{CH}+betavec{AH}tag{2}$$
From (1) and (2) you get that:
$$frac12vec{AH}-frac12vec{CH}=-alphavec{CH}+betavec{AH}$$
$$(frac12-beta)vec{AH}+(alpha-frac12)vec{CH}=0tag{3}$$
Vectors $vec{AH}$ and $vec{CH}$ are not parallel! If you multiply these vectors with some real factors different from zero, you still get vectors that are not parallel (vectors are just stretched). Sum of two vectors that are not parallel cannot be zero. So (3) can be valid only if both scalars are equal to zero:
$$frac12 - beta =0, alpha-frac12=0$$
...which simply means that $alpha=beta=frac12$. Consequentially:
$$vec{OM}=frac12vec{CH}$$
$$OM=frac12 CH$$
answered Jan 22 at 13:50
OldboyOldboy
8,62711036
$endgroup$
$begingroup$
Outstanding solution, Oldboy. But my question is that which term is better for AH and CH vector? Collinearity or obliqueness? I didn't understand and I need more clarification about what you said that if we multiply (3) vectors with some real factors different from zero, we still get vectors that aren't collinear. If you get time, please explain that.
$endgroup$
– Anirban Niloy
Jan 22 at 14:43
1
$begingroup$
I think that you are right. I have used "collinear" while in fact I meant "parallel". I'm going to correct that. Basically, if you have two vectors that are not parallel their sum cannot be zero! Multiply these two vectors with some scalars, it won't change their orientation in space. Non-parallel vectors will still be non-parallel and their sum cannot be zero.
$endgroup$
– Oldboy
Jan 22 at 14:54
add a comment |
$begingroup$
The most ellegant proof (at least in my opinion) is vector based:
Vectors $vec{CH}$ and $vec{OM}$ are parallel as well as vectors $vec{AH}$ and $vec{ON}$ and therefore:
$$vec{OM}=alphavec{CH}$$
$$vec{ON}=betavec{AH}$$
...for some real values of $alpha,beta$.
Segment MN connects midpoints of sides $AB$ and $BC$ and therefore:
$$vec{MN}=frac12vec{AC}=frac12(vec{AH}+vec{HC})=frac12vec{AH}-frac12vec{CH}tag{1}$$
On the other side it is obvious that:
$$vec{MN}=vec{MO}+vec{ON}=-vec{OM}+vec{ON}=-alphavec{CH}+betavec{AH}tag{2}$$
From (1) and (2) you get that:
$$frac12vec{AH}-frac12vec{CH}=-alphavec{CH}+betavec{AH}$$
$$(frac12-beta)vec{AH}+(alpha-frac12)vec{CH}=0tag{3}$$
Vectors $vec{AH}$ and $vec{CH}$ are not parallel! If you multiply these vectors with some real factors different from zero, you still get vectors that are not parallel (vectors are just stretched). Sum of two vectors that are not parallel cannot be zero. So (3) can be valid only if both scalars are equal to zero:
$$frac12 - beta =0, alpha-frac12=0$$
...which simply means that $alpha=beta=frac12$. Consequentially:
$$vec{OM}=frac12vec{CH}$$
$$OM=frac12 CH$$
answered Jan 22 at 13:50
OldboyOldboy
8,62711036
$endgroup$
$begingroup$
Outstanding solution, Oldboy. But my question is that which term is better for AH and CH vector? Collinearity or obliqueness? I didn't understand and I need more clarification about what you said that if we multiply (3) vectors with some real factors different from zero, we still get vectors that aren't collinear. If you get time, please explain that.
$endgroup$
– Anirban Niloy
Jan 22 at 14:43
1
$begingroup$
I think that you are right. I have used "collinear" while in fact I meant "parallel". I'm going to correct that. Basically, if you have two vectors that are not parallel their sum cannot be zero! Multiply these two vectors with some scalars, it won't change their orientation in space. Non-parallel vectors will still be non-parallel and their sum cannot be zero.
$endgroup$
– Oldboy
Jan 22 at 14:54
add a comment |
$begingroup$
The most ellegant proof (at least in my opinion) is vector based:
Vectors $vec{CH}$ and $vec{OM}$ are parallel as well as vectors $vec{AH}$ and $vec{ON}$ and therefore:
$$vec{OM}=alphavec{CH}$$
$$vec{ON}=betavec{AH}$$
...for some real values of $alpha,beta$.
Segment MN connects midpoints of sides $AB$ and $BC$ and therefore:
$$vec{MN}=frac12vec{AC}=frac12(vec{AH}+vec{HC})=frac12vec{AH}-frac12vec{CH}tag{1}$$
On the other side it is obvious that:
$$vec{MN}=vec{MO}+vec{ON}=-vec{OM}+vec{ON}=-alphavec{CH}+betavec{AH}tag{2}$$
From (1) and (2) you get that:
$$frac12vec{AH}-frac12vec{CH}=-alphavec{CH}+betavec{AH}$$
$$(frac12-beta)vec{AH}+(alpha-frac12)vec{CH}=0tag{3}$$
Vectors $vec{AH}$ and $vec{CH}$ are not parallel! If you multiply these vectors with some real factors different from zero, you still get vectors that are not parallel (vectors are just stretched). Sum of two vectors that are not parallel cannot be zero. So (3) can be valid only if both scalars are equal to zero:
$$frac12 - beta =0, alpha-frac12=0$$
...which simply means that $alpha=beta=frac12$. Consequentially:
$$vec{OM}=frac12vec{CH}$$
$$OM=frac12 CH$$
answered Jan 22 at 13:50
OldboyOldboy
8,62711036
$endgroup$
The most ellegant proof (at least in my opinion) is vector based:
Vectors $vec{CH}$ and $vec{OM}$ are parallel as well as vectors $vec{AH}$ and $vec{ON}$ and therefore:
$$vec{OM}=alphavec{CH}$$
$$vec{ON}=betavec{AH}$$
...for some real values of $alpha,beta$.
Segment MN connects midpoints of sides $AB$ and $BC$ and therefore:
$$vec{MN}=frac12vec{AC}=frac12(vec{AH}+vec{HC})=frac12vec{AH}-frac12vec{CH}tag{1}$$
On the other side it is obvious that:
$$vec{MN}=vec{MO}+vec{ON}=-vec{OM}+vec{ON}=-alphavec{CH}+betavec{AH}tag{2}$$
From (1) and (2) you get that:
$$frac12vec{AH}-frac12vec{CH}=-alphavec{CH}+betavec{AH}$$
$$(frac12-beta)vec{AH}+(alpha-frac12)vec{CH}=0tag{3}$$
Vectors $vec{AH}$ and $vec{CH}$ are not parallel! If you multiply these vectors with some real factors different from zero, you still get vectors that are not parallel (vectors are just stretched). Sum of two vectors that are not parallel cannot be zero. So (3) can be valid only if both scalars are equal to zero:
$$frac12 - beta =0, alpha-frac12=0$$
...which simply means that $alpha=beta=frac12$. Consequentially:
$$vec{OM}=frac12vec{CH}$$
$$OM=frac12 CH$$
answered Jan 22 at 13:50
OldboyOldboy
8,62711036
edited Jan 22 at 14:55
answered Jan 22 at 13:50
OldboyOldboy
8,62711036
answered Jan 22 at 13:50
OldboyOldboy
8,62711036
answered Jan 22 at 13:50
OldboyOldboy
8,62711036
8,62711036
$begingroup$
Outstanding solution, Oldboy. But my question is that which term is better for AH and CH vector? Collinearity or obliqueness? I didn't understand and I need more clarification about what you said that if we multiply (3) vectors with some real factors different from zero, we still get vectors that aren't collinear. If you get time, please explain that.
$endgroup$
– Anirban Niloy
Jan 22 at 14:43
1
$begingroup$
I think that you are right. I have used "collinear" while in fact I meant "parallel". I'm going to correct that. Basically, if you have two vectors that are not parallel their sum cannot be zero! Multiply these two vectors with some scalars, it won't change their orientation in space. Non-parallel vectors will still be non-parallel and their sum cannot be zero.
$endgroup$
– Oldboy
Jan 22 at 14:54
add a comment |
$begingroup$
Outstanding solution, Oldboy. But my question is that which term is better for AH and CH vector? Collinearity or obliqueness? I didn't understand and I need more clarification about what you said that if we multiply (3) vectors with some real factors different from zero, we still get vectors that aren't collinear. If you get time, please explain that.
$endgroup$
– Anirban Niloy
Jan 22 at 14:43
1
$begingroup$
I think that you are right. I have used "collinear" while in fact I meant "parallel". I'm going to correct that. Basically, if you have two vectors that are not parallel their sum cannot be zero! Multiply these two vectors with some scalars, it won't change their orientation in space. Non-parallel vectors will still be non-parallel and their sum cannot be zero.
$endgroup$
– Oldboy
Jan 22 at 14:54
$begingroup$
Outstanding solution, Oldboy. But my question is that which term is better for AH and CH vector? Collinearity or obliqueness? I didn't understand and I need more clarification about what you said that if we multiply (3) vectors with some real factors different from zero, we still get vectors that aren't collinear. If you get time, please explain that.
$endgroup$
– Anirban Niloy
Jan 22 at 14:43
$begingroup$
Outstanding solution, Oldboy. But my question is that which term is better for AH and CH vector? Collinearity or obliqueness? I didn't understand and I need more clarification about what you said that if we multiply (3) vectors with some real factors different from zero, we still get vectors that aren't collinear. If you get time, please explain that.
$endgroup$
– Anirban Niloy
Jan 22 at 14:43
1
1
$begingroup$
I think that you are right. I have used "collinear" while in fact I meant "parallel". I'm going to correct that. Basically, if you have two vectors that are not parallel their sum cannot be zero! Multiply these two vectors with some scalars, it won't change their orientation in space. Non-parallel vectors will still be non-parallel and their sum cannot be zero.
$endgroup$
– Oldboy
Jan 22 at 14:54
$begingroup$
I think that you are right. I have used "collinear" while in fact I meant "parallel". I'm going to correct that. Basically, if you have two vectors that are not parallel their sum cannot be zero! Multiply these two vectors with some scalars, it won't change their orientation in space. Non-parallel vectors will still be non-parallel and their sum cannot be zero.
$endgroup$
– Oldboy
Jan 22 at 14:54
add a comment |
$begingroup$
Do you know Hamilton theorem? It says $$vec{OH}= vec{OA}+vec{OB}+vec{OC} $$
So $$vec{CH}=vec{OH}- vec{OC}= vec{OA}+vec{OB} ={1over 2}vec{OM} $$
answered Feb 4 at 16:58
Maria MazurMaria Mazur
46.4k1160119
$endgroup$
$begingroup$
No, I didn't know about it. You helped me to learn a new thing which is more appreciated. If possible, I will try to read about the personal details of Hamilton theorem.
$endgroup$
– Anirban Niloy
Feb 5 at 1:38
add a comment |
$begingroup$
Do you know Hamilton theorem? It says $$vec{OH}= vec{OA}+vec{OB}+vec{OC} $$
So $$vec{CH}=vec{OH}- vec{OC}= vec{OA}+vec{OB} ={1over 2}vec{OM} $$
answered Feb 4 at 16:58
Maria MazurMaria Mazur
46.4k1160119
$endgroup$
$begingroup$
No, I didn't know about it. You helped me to learn a new thing which is more appreciated. If possible, I will try to read about the personal details of Hamilton theorem.
$endgroup$
– Anirban Niloy
Feb 5 at 1:38
add a comment |
$begingroup$
Do you know Hamilton theorem? It says $$vec{OH}= vec{OA}+vec{OB}+vec{OC} $$
So $$vec{CH}=vec{OH}- vec{OC}= vec{OA}+vec{OB} ={1over 2}vec{OM} $$
answered Feb 4 at 16:58
Maria MazurMaria Mazur
46.4k1160119
$endgroup$
Do you know Hamilton theorem? It says $$vec{OH}= vec{OA}+vec{OB}+vec{OC} $$
So $$vec{CH}=vec{OH}- vec{OC}= vec{OA}+vec{OB} ={1over 2}vec{OM} $$
answered Feb 4 at 16:58
Maria MazurMaria Mazur
46.4k1160119
answered Feb 4 at 16:58
Maria MazurMaria Mazur
46.4k1160119
answered Feb 4 at 16:58
Maria MazurMaria Mazur
46.4k1160119
answered Feb 4 at 16:58
Maria MazurMaria Mazur
46.4k1160119
46.4k1160119
$begingroup$
No, I didn't know about it. You helped me to learn a new thing which is more appreciated. If possible, I will try to read about the personal details of Hamilton theorem.
$endgroup$
– Anirban Niloy
Feb 5 at 1:38
add a comment |
$begingroup$
No, I didn't know about it. You helped me to learn a new thing which is more appreciated. If possible, I will try to read about the personal details of Hamilton theorem.
$endgroup$
– Anirban Niloy
Feb 5 at 1:38
$begingroup$
No, I didn't know about it. You helped me to learn a new thing which is more appreciated. If possible, I will try to read about the personal details of Hamilton theorem.
$endgroup$
– Anirban Niloy
Feb 5 at 1:38
$begingroup$
No, I didn't know about it. You helped me to learn a new thing which is more appreciated. If possible, I will try to read about the personal details of Hamilton theorem.
$endgroup$
– Anirban Niloy
Feb 5 at 1:38
add a comment |
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$begingroup$
If $CH=2OM$ was proved while proving another equality, then you already have an independent proof of that.
$endgroup$
– Aretino
Jan 22 at 14:14
$begingroup$
I think, I have been futile to clearify my text. While proving another equality, I just used CH = 2OM as a preknowledge, not was proved in that proof. So I wanted a distinctive proof of above condition which has been proposed.
$endgroup$
– Anirban Niloy
Jan 22 at 14:22