Proof - Sum of Two sides of Triangle is greater than twice of its median. [closed]

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Proof that the sum of any two sides of triangle is always greater than twice of its median.Without using any type of Construction

inequality triangle alternative-proof plane-geometry

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edited Jul 17 at 2:53

asked Jul 17 at 1:44

Ashutosh Kumar

5710

closed as off-topic by Alex Francisco, Trần Thúc Minh Trí, max_zorn, Isaac Browne, Claude Leibovici Jul 17 at 9:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Trần Thúc Minh Trí, max_zorn, Isaac Browne, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hint: complete the parallelogram $,ABcolorredA'C,$, then use that $,AB+BA' gt AA',$.
  – dxiv
  Jul 17 at 1:53
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Sum of Two sides of triangle is greater than twice of its median
  – tien lee
  Jul 17 at 3:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee This one was posted first. You really should mark the other one as a duplicate.
  – dxiv
  Jul 17 at 3:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @dxiv In fact both these questions should be closed. The reason why he posted the same question again might be that he did not get the answer which he want.
  – tien lee
  Jul 17 at 3:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee Then downvote it, or vote to close as PSQ, rather than flagging it as a duplicate of a later question (which the OP deleted since). As for did not get the answer which he want at least one of the answers below uses no "type of construction", which would seemingly fit OP's bill. But then I agree, the question is completely missing any context.
  – dxiv
  Jul 17 at 3:12
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

2

Proof that the sum of any two sides of triangle is always greater than twice of its median.Without using any type of Construction

inequality triangle alternative-proof plane-geometry

share|cite|improve this question

edited Jul 17 at 2:53

asked Jul 17 at 1:44

Ashutosh Kumar

5710

closed as off-topic by Alex Francisco, Trần Thúc Minh Trí, max_zorn, Isaac Browne, Claude Leibovici Jul 17 at 9:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Trần Thúc Minh Trí, max_zorn, Isaac Browne, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hint: complete the parallelogram $,ABcolorredA'C,$, then use that $,AB+BA' gt AA',$.
  – dxiv
  Jul 17 at 1:53
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Sum of Two sides of triangle is greater than twice of its median
  – tien lee
  Jul 17 at 3:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee This one was posted first. You really should mark the other one as a duplicate.
  – dxiv
  Jul 17 at 3:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @dxiv In fact both these questions should be closed. The reason why he posted the same question again might be that he did not get the answer which he want.
  – tien lee
  Jul 17 at 3:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee Then downvote it, or vote to close as PSQ, rather than flagging it as a duplicate of a later question (which the OP deleted since). As for did not get the answer which he want at least one of the answers below uses no "type of construction", which would seemingly fit OP's bill. But then I agree, the question is completely missing any context.
  – dxiv
  Jul 17 at 3:12
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

2

up vote
0
down vote

favorite

2

2

Proof that the sum of any two sides of triangle is always greater than twice of its median.Without using any type of Construction

inequality triangle alternative-proof plane-geometry

share|cite|improve this question

edited Jul 17 at 2:53

asked Jul 17 at 1:44

Ashutosh Kumar

5710

Proof that the sum of any two sides of triangle is always greater than twice of its median.Without using any type of Construction

inequality triangle alternative-proof plane-geometry

share|cite|improve this question

edited Jul 17 at 2:53

asked Jul 17 at 1:44

Ashutosh Kumar

5710

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 2:53

edited Jul 17 at 2:53

edited Jul 17 at 2:53

asked Jul 17 at 1:44

Ashutosh Kumar

5710

asked Jul 17 at 1:44

Ashutosh Kumar

5710

asked Jul 17 at 1:44

Ashutosh Kumar

5710

5710

closed as off-topic by Alex Francisco, Trần Thúc Minh Trí, max_zorn, Isaac Browne, Claude Leibovici Jul 17 at 9:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Trần Thúc Minh Trí, max_zorn, Isaac Browne, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Alex Francisco, Trần Thúc Minh Trí, max_zorn, Isaac Browne, Claude Leibovici Jul 17 at 9:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Trần Thúc Minh Trí, max_zorn, Isaac Browne, Claude Leibovici

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hint: complete the parallelogram $,ABcolorredA'C,$, then use that $,AB+BA' gt AA',$.
  – dxiv
  Jul 17 at 1:53
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Sum of Two sides of triangle is greater than twice of its median
  – tien lee
  Jul 17 at 3:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee This one was posted first. You really should mark the other one as a duplicate.
  – dxiv
  Jul 17 at 3:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @dxiv In fact both these questions should be closed. The reason why he posted the same question again might be that he did not get the answer which he want.
  – tien lee
  Jul 17 at 3:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee Then downvote it, or vote to close as PSQ, rather than flagging it as a duplicate of a later question (which the OP deleted since). As for did not get the answer which he want at least one of the answers below uses no "type of construction", which would seemingly fit OP's bill. But then I agree, the question is completely missing any context.
  – dxiv
  Jul 17 at 3:12
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hint: complete the parallelogram $,ABcolorredA'C,$, then use that $,AB+BA' gt AA',$.
  – dxiv
  Jul 17 at 1:53
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Sum of Two sides of triangle is greater than twice of its median
  – tien lee
  Jul 17 at 3:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee This one was posted first. You really should mark the other one as a duplicate.
  – dxiv
  Jul 17 at 3:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @dxiv In fact both these questions should be closed. The reason why he posted the same question again might be that he did not get the answer which he want.
  – tien lee
  Jul 17 at 3:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee Then downvote it, or vote to close as PSQ, rather than flagging it as a duplicate of a later question (which the OP deleted since). As for did not get the answer which he want at least one of the answers below uses no "type of construction", which would seemingly fit OP's bill. But then I agree, the question is completely missing any context.
  – dxiv
  Jul 17 at 3:12
  
  

  
  
  
  

1

1

Hint: complete the parallelogram $,ABcolorredA'C,$, then use that $,AB+BA' gt AA',$.
– dxiv
Jul 17 at 1:53

Hint: complete the parallelogram $,ABcolorredA'C,$, then use that $,AB+BA' gt AA',$.
– dxiv
Jul 17 at 1:53

1

1

Possible duplicate of Sum of Two sides of triangle is greater than twice of its median
– tien lee
Jul 17 at 3:02

Possible duplicate of Sum of Two sides of triangle is greater than twice of its median
– tien lee
Jul 17 at 3:02

@tienlee This one was posted first. You really should mark the other one as a duplicate.
– dxiv
Jul 17 at 3:02

@tienlee This one was posted first. You really should mark the other one as a duplicate.
– dxiv
Jul 17 at 3:02

2

2

@dxiv In fact both these questions should be closed. The reason why he posted the same question again might be that he did not get the answer which he want.
– tien lee
Jul 17 at 3:04

@dxiv In fact both these questions should be closed. The reason why he posted the same question again might be that he did not get the answer which he want.
– tien lee
Jul 17 at 3:04

@tienlee Then downvote it, or vote to close as PSQ, rather than flagging it as a duplicate of a later question (which the OP deleted since). As for did not get the answer which he want at least one of the answers below uses no "type of construction", which would seemingly fit OP's bill. But then I agree, the question is completely missing any context.
– dxiv
Jul 17 at 3:12

@tienlee Then downvote it, or vote to close as PSQ, rather than flagging it as a duplicate of a later question (which the OP deleted since). As for did not get the answer which he want at least one of the answers below uses no "type of construction", which would seemingly fit OP's bill. But then I agree, the question is completely missing any context.
– dxiv
Jul 17 at 3:12

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote

Take $E$ the middle of $AC$. Then one has:

$$AD<DE+AE=fracAB+AC2$$

share|cite|improve this answer

answered Jul 17 at 1:48

Momo

11.9k21330

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the answer is downvoted?
  – Momo
  Jul 17 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wasn't $D$ already defined to be the "middle" of $AC text ? qquad$
  – Michael Hardy
  Jul 17 at 3:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E already defined to be the "middle" of AC
  – Ashutosh Kumar
  Jul 17 at 3:09
  

  
  
  
  

add a comment | 

up vote
1
down vote

According to the triangle inequality, $$ |a+b|le |a|+|b| $$

Note that if sides of the triangle of vectors $vec a$ and $vec b$ then the median is $$frac vec a+b2$$

share|cite|improve this answer

edited Jul 17 at 3:06

Michael Hardy

204k23186462

answered Jul 17 at 2:06

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

As @dxiv said in his comment, make a copy of the triangle and have the lines $BC$ on both triangles connected. but flip the copy of the triangle such that lines $AB$ and $A'C'$ are on the same side, thus giving you a parallelogram. Now the medians of both triangles from $AD$ and $A'D$ form a line, which is twice the median. So now apply the triangle inequality to get $AB + BA' > ADA'$, where $BA'equiv AC$ and $ADA'$ is twice the median

share|cite|improve this answer

edited Jul 17 at 2:10

answered Jul 17 at 2:02

john fowles

1,093817

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Take $E$ the middle of $AC$. Then one has:

$$AD<DE+AE=fracAB+AC2$$

share|cite|improve this answer

answered Jul 17 at 1:48

Momo

11.9k21330

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the answer is downvoted?
  – Momo
  Jul 17 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wasn't $D$ already defined to be the "middle" of $AC text ? qquad$
  – Michael Hardy
  Jul 17 at 3:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E already defined to be the "middle" of AC
  – Ashutosh Kumar
  Jul 17 at 3:09
  

  
  
  
  

add a comment | 

up vote
1
down vote

Take $E$ the middle of $AC$. Then one has:

$$AD<DE+AE=fracAB+AC2$$

share|cite|improve this answer

answered Jul 17 at 1:48

Momo

11.9k21330

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the answer is downvoted?
  – Momo
  Jul 17 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wasn't $D$ already defined to be the "middle" of $AC text ? qquad$
  – Michael Hardy
  Jul 17 at 3:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E already defined to be the "middle" of AC
  – Ashutosh Kumar
  Jul 17 at 3:09
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Take $E$ the middle of $AC$. Then one has:

$$AD<DE+AE=fracAB+AC2$$

share|cite|improve this answer

answered Jul 17 at 1:48

Momo

11.9k21330

Take $E$ the middle of $AC$. Then one has:

$$AD<DE+AE=fracAB+AC2$$

share|cite|improve this answer

answered Jul 17 at 1:48

Momo

11.9k21330

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 1:48

Momo

11.9k21330

answered Jul 17 at 1:48

Momo

11.9k21330

answered Jul 17 at 1:48

Momo

11.9k21330

11.9k21330

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the answer is downvoted?
  – Momo
  Jul 17 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wasn't $D$ already defined to be the "middle" of $AC text ? qquad$
  – Michael Hardy
  Jul 17 at 3:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E already defined to be the "middle" of AC
  – Ashutosh Kumar
  Jul 17 at 3:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the answer is downvoted?
  – Momo
  Jul 17 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wasn't $D$ already defined to be the "middle" of $AC text ? qquad$
  – Michael Hardy
  Jul 17 at 3:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  E already defined to be the "middle" of AC
  – Ashutosh Kumar
  Jul 17 at 3:09
  

  
  
  
  

Why the answer is downvoted?
– Momo
Jul 17 at 2:03

Why the answer is downvoted?
– Momo
Jul 17 at 2:03

Wasn't $D$ already defined to be the "middle" of $AC text ? qquad$
– Michael Hardy
Jul 17 at 3:05

Wasn't $D$ already defined to be the "middle" of $AC text ? qquad$
– Michael Hardy
Jul 17 at 3:05

E already defined to be the "middle" of AC
– Ashutosh Kumar
Jul 17 at 3:09

E already defined to be the "middle" of AC
– Ashutosh Kumar
Jul 17 at 3:09

add a comment | 

up vote
1
down vote

According to the triangle inequality, $$ |a+b|le |a|+|b| $$

Note that if sides of the triangle of vectors $vec a$ and $vec b$ then the median is $$frac vec a+b2$$

share|cite|improve this answer

edited Jul 17 at 3:06

Michael Hardy

204k23186462

answered Jul 17 at 2:06

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
1
down vote

According to the triangle inequality, $$ |a+b|le |a|+|b| $$

Note that if sides of the triangle of vectors $vec a$ and $vec b$ then the median is $$frac vec a+b2$$

share|cite|improve this answer

edited Jul 17 at 3:06

Michael Hardy

204k23186462

answered Jul 17 at 2:06

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
1
down vote

up vote
1
down vote

According to the triangle inequality, $$ |a+b|le |a|+|b| $$

Note that if sides of the triangle of vectors $vec a$ and $vec b$ then the median is $$frac vec a+b2$$

share|cite|improve this answer

edited Jul 17 at 3:06

Michael Hardy

204k23186462

answered Jul 17 at 2:06

Mohammad Riazi-Kermani

27.5k41852

According to the triangle inequality, $$ |a+b|le |a|+|b| $$

Note that if sides of the triangle of vectors $vec a$ and $vec b$ then the median is $$frac vec a+b2$$

share|cite|improve this answer

edited Jul 17 at 3:06

Michael Hardy

204k23186462

answered Jul 17 at 2:06

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 3:06

Michael Hardy

204k23186462

edited Jul 17 at 3:06

Michael Hardy

204k23186462

edited Jul 17 at 3:06

Michael Hardy

204k23186462

204k23186462

answered Jul 17 at 2:06

Mohammad Riazi-Kermani

27.5k41852

answered Jul 17 at 2:06

Mohammad Riazi-Kermani

27.5k41852

answered Jul 17 at 2:06

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

add a comment | 

add a comment | 

up vote
0
down vote

As @dxiv said in his comment, make a copy of the triangle and have the lines $BC$ on both triangles connected. but flip the copy of the triangle such that lines $AB$ and $A'C'$ are on the same side, thus giving you a parallelogram. Now the medians of both triangles from $AD$ and $A'D$ form a line, which is twice the median. So now apply the triangle inequality to get $AB + BA' > ADA'$, where $BA'equiv AC$ and $ADA'$ is twice the median

share|cite|improve this answer

edited Jul 17 at 2:10

answered Jul 17 at 2:02

john fowles

1,093817

add a comment | 

up vote
0
down vote

As @dxiv said in his comment, make a copy of the triangle and have the lines $BC$ on both triangles connected. but flip the copy of the triangle such that lines $AB$ and $A'C'$ are on the same side, thus giving you a parallelogram. Now the medians of both triangles from $AD$ and $A'D$ form a line, which is twice the median. So now apply the triangle inequality to get $AB + BA' > ADA'$, where $BA'equiv AC$ and $ADA'$ is twice the median

share|cite|improve this answer

edited Jul 17 at 2:10

answered Jul 17 at 2:02

john fowles

1,093817

add a comment | 

up vote
0
down vote

up vote
0
down vote

As @dxiv said in his comment, make a copy of the triangle and have the lines $BC$ on both triangles connected. but flip the copy of the triangle such that lines $AB$ and $A'C'$ are on the same side, thus giving you a parallelogram. Now the medians of both triangles from $AD$ and $A'D$ form a line, which is twice the median. So now apply the triangle inequality to get $AB + BA' > ADA'$, where $BA'equiv AC$ and $ADA'$ is twice the median

share|cite|improve this answer

edited Jul 17 at 2:10

answered Jul 17 at 2:02

john fowles

1,093817

As @dxiv said in his comment, make a copy of the triangle and have the lines $BC$ on both triangles connected. but flip the copy of the triangle such that lines $AB$ and $A'C'$ are on the same side, thus giving you a parallelogram. Now the medians of both triangles from $AD$ and $A'D$ form a line, which is twice the median. So now apply the triangle inequality to get $AB + BA' > ADA'$, where $BA'equiv AC$ and $ADA'$ is twice the median

share|cite|improve this answer

edited Jul 17 at 2:10

answered Jul 17 at 2:02

john fowles

1,093817

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 2:10

edited Jul 17 at 2:10

edited Jul 17 at 2:10

answered Jul 17 at 2:02

john fowles

1,093817

answered Jul 17 at 2:02

john fowles

1,093817

answered Jul 17 at 2:02

john fowles

1,093817

1,093817

add a comment | 

add a comment |