Proof on triangles [closed]

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I've already seen that this is true visually(I made a sketch), but I don't know how to write it formally to prove it...I'm studying geometry formally for the first time.

QUESTION: If $overline AB cong overline DE,overline AC cong overline DF, angle CBA cong angle FED$, but $triangle ABC$ and $triangle DEF$ aren't congruent, prove that if $|CB|lt|EF|$ then $angle BCA gt 90$.

Any tips?

geometry euclidean-geometry triangle

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edited Jul 28 at 20:48

asked Jul 28 at 19:36

Pedro Fernandes

186

closed as off-topic by Did, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Key Flex Jul 29 at 0:04

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." – Did, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Key Flex

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you post the sketch?
  – John Glenn
  Jul 28 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnGlenn I can't right now. I drew triangle EDF(base EF) with angle FED being an acute angle. Then I drew triangle BAC(base BC) and I rotated the side AC to the left, until |BC|<|EF| and angle BCA is obviously obtuse.
  – Pedro Fernandes
  Jul 28 at 19:48
  

  
  
  
  

add a comment | 

up vote
-4
down vote

favorite

I've already seen that this is true visually(I made a sketch), but I don't know how to write it formally to prove it...I'm studying geometry formally for the first time.

QUESTION: If $overline AB cong overline DE,overline AC cong overline DF, angle CBA cong angle FED$, but $triangle ABC$ and $triangle DEF$ aren't congruent, prove that if $|CB|lt|EF|$ then $angle BCA gt 90$.

Any tips?

geometry euclidean-geometry triangle

share|cite|improve this question

edited Jul 28 at 20:48

asked Jul 28 at 19:36

Pedro Fernandes

186

closed as off-topic by Did, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Key Flex Jul 29 at 0:04

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." – Did, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Key Flex

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you post the sketch?
  – John Glenn
  Jul 28 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnGlenn I can't right now. I drew triangle EDF(base EF) with angle FED being an acute angle. Then I drew triangle BAC(base BC) and I rotated the side AC to the left, until |BC|<|EF| and angle BCA is obviously obtuse.
  – Pedro Fernandes
  Jul 28 at 19:48
  

  
  
  
  

add a comment | 

up vote
-4
down vote

favorite

up vote
-4
down vote

favorite

I've already seen that this is true visually(I made a sketch), but I don't know how to write it formally to prove it...I'm studying geometry formally for the first time.

QUESTION: If $overline AB cong overline DE,overline AC cong overline DF, angle CBA cong angle FED$, but $triangle ABC$ and $triangle DEF$ aren't congruent, prove that if $|CB|lt|EF|$ then $angle BCA gt 90$.

Any tips?

geometry euclidean-geometry triangle

share|cite|improve this question

edited Jul 28 at 20:48

asked Jul 28 at 19:36

Pedro Fernandes

186

I've already seen that this is true visually(I made a sketch), but I don't know how to write it formally to prove it...I'm studying geometry formally for the first time.

QUESTION: If $overline AB cong overline DE,overline AC cong overline DF, angle CBA cong angle FED$, but $triangle ABC$ and $triangle DEF$ aren't congruent, prove that if $|CB|lt|EF|$ then $angle BCA gt 90$.

Any tips?

geometry euclidean-geometry triangle

share|cite|improve this question

edited Jul 28 at 20:48

asked Jul 28 at 19:36

Pedro Fernandes

186

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 20:48

edited Jul 28 at 20:48

edited Jul 28 at 20:48

asked Jul 28 at 19:36

Pedro Fernandes

186

asked Jul 28 at 19:36

Pedro Fernandes

186

asked Jul 28 at 19:36

Pedro Fernandes

186

186

closed as off-topic by Did, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Key Flex Jul 29 at 0:04

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." – Did, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Key Flex

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

closed as off-topic by Did, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Key Flex Jul 29 at 0:04

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." – Did, Jyrki Lahtonen, Xander Henderson, José Carlos Santos, Key Flex

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you post the sketch?
  – John Glenn
  Jul 28 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnGlenn I can't right now. I drew triangle EDF(base EF) with angle FED being an acute angle. Then I drew triangle BAC(base BC) and I rotated the side AC to the left, until |BC|<|EF| and angle BCA is obviously obtuse.
  – Pedro Fernandes
  Jul 28 at 19:48
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you post the sketch?
  – John Glenn
  Jul 28 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JohnGlenn I can't right now. I drew triangle EDF(base EF) with angle FED being an acute angle. Then I drew triangle BAC(base BC) and I rotated the side AC to the left, until |BC|<|EF| and angle BCA is obviously obtuse.
  – Pedro Fernandes
  Jul 28 at 19:48
  

  
  
  
  

Can you post the sketch?
– John Glenn
Jul 28 at 19:43

Can you post the sketch?
– John Glenn
Jul 28 at 19:43

@JohnGlenn I can't right now. I drew triangle EDF(base EF) with angle FED being an acute angle. Then I drew triangle BAC(base BC) and I rotated the side AC to the left, until |BC|<|EF| and angle BCA is obviously obtuse.
– Pedro Fernandes
Jul 28 at 19:48

@JohnGlenn I can't right now. I drew triangle EDF(base EF) with angle FED being an acute angle. Then I drew triangle BAC(base BC) and I rotated the side AC to the left, until |BC|<|EF| and angle BCA is obviously obtuse.
– Pedro Fernandes
Jul 28 at 19:48

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote

Draw triangles $ABC,DEF$. Locate C' on EF such that C'E=BC. Now prove that triangles $ABC,DEC'$ are congruent and then solve the rest.

share|cite|improve this answer

answered Jul 28 at 20:02

Love Invariants

77715

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This was useful, but it's not what I'm looking for. I haven't yet reached the theorem that says that the angles of a triangle sum to 180. I only proved that the sum of the angles of a triangle are at most 180. But I would be happy with this proof when I reach there, but there's one thing that is bugging me, how do you justify that a<90? Just by looking at the triangle?
  – Pedro Fernandes
  Jul 28 at 20:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  . In $triangle DC′F$ we have 2 angles which are α which means maximum value of α is $90^∘$. Observe that maximum sum of angles in a triangle can be $180^circ$ then sum of all angles in $C'FD$ is between $0$ and $180$ which itself proves that $alpha$ is atleast $90^circ$
  – Love Invariants
  Jul 28 at 21:59
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Notice that as $EF > BC$, exists $G$ such that $m(EG)=m(BC)$.

Then, by SAS, $ABC cong DEG$ and we have that $m(DG)=m(AC)$, and the angles $BCA cong DGE$

Now, the triangle $DGF$ is isosceles, then $DGF cong DFG$.

If, $BCA le 90$ follows that $DGF ge 90$ because they are supplementary.

And we have a triangle $DGF$ with two obtuseright angles, a contradicition. (Why?)

Notice that this result is true in any Neutral Geometry

share|cite|improve this answer

answered Jul 28 at 22:50

idk

9431414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think your 2nd figure is wrong, but your answer gave me an idea and I managed to prove this! I will post it.
  – Pedro Fernandes
  Jul 28 at 23:12
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

I managed to solve this! I came up with this drawing when I saw @idk answer.
Since $triangle ACF$ is isosceles we have that $angle FCA =angle CFA$.
If $angle BCA=90$ then $angle FCA=90,$ but then $angle CFA=90$, which is a contradiction since the sum of the angles of a triangle is $leq180$. If $angle BCA<90$ then $angle FCAgt90$, but then $angle CFAgt90$, which is a contradiction for the same reason. In conclusion it must be $angle BCAgt90.$

share|cite|improve this answer

edited Jul 28 at 23:49

answered Jul 28 at 23:41

Pedro Fernandes

186

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

Draw triangles $ABC,DEF$. Locate C' on EF such that C'E=BC. Now prove that triangles $ABC,DEC'$ are congruent and then solve the rest.

share|cite|improve this answer

answered Jul 28 at 20:02

Love Invariants

77715

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This was useful, but it's not what I'm looking for. I haven't yet reached the theorem that says that the angles of a triangle sum to 180. I only proved that the sum of the angles of a triangle are at most 180. But I would be happy with this proof when I reach there, but there's one thing that is bugging me, how do you justify that a<90? Just by looking at the triangle?
  – Pedro Fernandes
  Jul 28 at 20:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  . In $triangle DC′F$ we have 2 angles which are α which means maximum value of α is $90^∘$. Observe that maximum sum of angles in a triangle can be $180^circ$ then sum of all angles in $C'FD$ is between $0$ and $180$ which itself proves that $alpha$ is atleast $90^circ$
  – Love Invariants
  Jul 28 at 21:59
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Draw triangles $ABC,DEF$. Locate C' on EF such that C'E=BC. Now prove that triangles $ABC,DEC'$ are congruent and then solve the rest.

share|cite|improve this answer

answered Jul 28 at 20:02

Love Invariants

77715

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This was useful, but it's not what I'm looking for. I haven't yet reached the theorem that says that the angles of a triangle sum to 180. I only proved that the sum of the angles of a triangle are at most 180. But I would be happy with this proof when I reach there, but there's one thing that is bugging me, how do you justify that a<90? Just by looking at the triangle?
  – Pedro Fernandes
  Jul 28 at 20:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  . In $triangle DC′F$ we have 2 angles which are α which means maximum value of α is $90^∘$. Observe that maximum sum of angles in a triangle can be $180^circ$ then sum of all angles in $C'FD$ is between $0$ and $180$ which itself proves that $alpha$ is atleast $90^circ$
  – Love Invariants
  Jul 28 at 21:59
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Draw triangles $ABC,DEF$. Locate C' on EF such that C'E=BC. Now prove that triangles $ABC,DEC'$ are congruent and then solve the rest.

share|cite|improve this answer

answered Jul 28 at 20:02

Love Invariants

77715

Draw triangles $ABC,DEF$. Locate C' on EF such that C'E=BC. Now prove that triangles $ABC,DEC'$ are congruent and then solve the rest.

share|cite|improve this answer

answered Jul 28 at 20:02

Love Invariants

77715

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 20:02

Love Invariants

77715

answered Jul 28 at 20:02

Love Invariants

77715

answered Jul 28 at 20:02

Love Invariants

77715

77715

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This was useful, but it's not what I'm looking for. I haven't yet reached the theorem that says that the angles of a triangle sum to 180. I only proved that the sum of the angles of a triangle are at most 180. But I would be happy with this proof when I reach there, but there's one thing that is bugging me, how do you justify that a<90? Just by looking at the triangle?
  – Pedro Fernandes
  Jul 28 at 20:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  . In $triangle DC′F$ we have 2 angles which are α which means maximum value of α is $90^∘$. Observe that maximum sum of angles in a triangle can be $180^circ$ then sum of all angles in $C'FD$ is between $0$ and $180$ which itself proves that $alpha$ is atleast $90^circ$
  – Love Invariants
  Jul 28 at 21:59
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This was useful, but it's not what I'm looking for. I haven't yet reached the theorem that says that the angles of a triangle sum to 180. I only proved that the sum of the angles of a triangle are at most 180. But I would be happy with this proof when I reach there, but there's one thing that is bugging me, how do you justify that a<90? Just by looking at the triangle?
  – Pedro Fernandes
  Jul 28 at 20:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  . In $triangle DC′F$ we have 2 angles which are α which means maximum value of α is $90^∘$. Observe that maximum sum of angles in a triangle can be $180^circ$ then sum of all angles in $C'FD$ is between $0$ and $180$ which itself proves that $alpha$ is atleast $90^circ$
  – Love Invariants
  Jul 28 at 21:59
  
  

  
  
  
  

This was useful, but it's not what I'm looking for. I haven't yet reached the theorem that says that the angles of a triangle sum to 180. I only proved that the sum of the angles of a triangle are at most 180. But I would be happy with this proof when I reach there, but there's one thing that is bugging me, how do you justify that a<90? Just by looking at the triangle?
– Pedro Fernandes
Jul 28 at 20:45

This was useful, but it's not what I'm looking for. I haven't yet reached the theorem that says that the angles of a triangle sum to 180. I only proved that the sum of the angles of a triangle are at most 180. But I would be happy with this proof when I reach there, but there's one thing that is bugging me, how do you justify that a<90? Just by looking at the triangle?
– Pedro Fernandes
Jul 28 at 20:45

. In $triangle DC′F$ we have 2 angles which are α which means maximum value of α is $90^∘$. Observe that maximum sum of angles in a triangle can be $180^circ$ then sum of all angles in $C'FD$ is between $0$ and $180$ which itself proves that $alpha$ is atleast $90^circ$
– Love Invariants
Jul 28 at 21:59

. In $triangle DC′F$ we have 2 angles which are α which means maximum value of α is $90^∘$. Observe that maximum sum of angles in a triangle can be $180^circ$ then sum of all angles in $C'FD$ is between $0$ and $180$ which itself proves that $alpha$ is atleast $90^circ$
– Love Invariants
Jul 28 at 21:59

add a comment | 

up vote
1
down vote

Notice that as $EF > BC$, exists $G$ such that $m(EG)=m(BC)$.

Then, by SAS, $ABC cong DEG$ and we have that $m(DG)=m(AC)$, and the angles $BCA cong DGE$

Now, the triangle $DGF$ is isosceles, then $DGF cong DFG$.

If, $BCA le 90$ follows that $DGF ge 90$ because they are supplementary.

And we have a triangle $DGF$ with two obtuseright angles, a contradicition. (Why?)

Notice that this result is true in any Neutral Geometry

share|cite|improve this answer

answered Jul 28 at 22:50

idk

9431414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think your 2nd figure is wrong, but your answer gave me an idea and I managed to prove this! I will post it.
  – Pedro Fernandes
  Jul 28 at 23:12
  

  
  
  
  

add a comment | 

up vote
1
down vote

Notice that as $EF > BC$, exists $G$ such that $m(EG)=m(BC)$.

Then, by SAS, $ABC cong DEG$ and we have that $m(DG)=m(AC)$, and the angles $BCA cong DGE$

Now, the triangle $DGF$ is isosceles, then $DGF cong DFG$.

If, $BCA le 90$ follows that $DGF ge 90$ because they are supplementary.

And we have a triangle $DGF$ with two obtuseright angles, a contradicition. (Why?)

Notice that this result is true in any Neutral Geometry

share|cite|improve this answer

answered Jul 28 at 22:50

idk

9431414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think your 2nd figure is wrong, but your answer gave me an idea and I managed to prove this! I will post it.
  – Pedro Fernandes
  Jul 28 at 23:12
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Notice that as $EF > BC$, exists $G$ such that $m(EG)=m(BC)$.

Then, by SAS, $ABC cong DEG$ and we have that $m(DG)=m(AC)$, and the angles $BCA cong DGE$

Now, the triangle $DGF$ is isosceles, then $DGF cong DFG$.

If, $BCA le 90$ follows that $DGF ge 90$ because they are supplementary.

And we have a triangle $DGF$ with two obtuseright angles, a contradicition. (Why?)

Notice that this result is true in any Neutral Geometry

share|cite|improve this answer

answered Jul 28 at 22:50

idk

9431414

Notice that as $EF > BC$, exists $G$ such that $m(EG)=m(BC)$.

Then, by SAS, $ABC cong DEG$ and we have that $m(DG)=m(AC)$, and the angles $BCA cong DGE$

Now, the triangle $DGF$ is isosceles, then $DGF cong DFG$.

If, $BCA le 90$ follows that $DGF ge 90$ because they are supplementary.

And we have a triangle $DGF$ with two obtuseright angles, a contradicition. (Why?)

Notice that this result is true in any Neutral Geometry

share|cite|improve this answer

answered Jul 28 at 22:50

idk

9431414

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 22:50

idk

9431414

answered Jul 28 at 22:50

idk

9431414

answered Jul 28 at 22:50

idk

9431414

9431414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think your 2nd figure is wrong, but your answer gave me an idea and I managed to prove this! I will post it.
  – Pedro Fernandes
  Jul 28 at 23:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think your 2nd figure is wrong, but your answer gave me an idea and I managed to prove this! I will post it.
  – Pedro Fernandes
  Jul 28 at 23:12
  

  
  
  
  

I think your 2nd figure is wrong, but your answer gave me an idea and I managed to prove this! I will post it.
– Pedro Fernandes
Jul 28 at 23:12

I think your 2nd figure is wrong, but your answer gave me an idea and I managed to prove this! I will post it.
– Pedro Fernandes
Jul 28 at 23:12

add a comment | 

up vote
0
down vote



accepted

I managed to solve this! I came up with this drawing when I saw @idk answer.
Since $triangle ACF$ is isosceles we have that $angle FCA =angle CFA$.
If $angle BCA=90$ then $angle FCA=90,$ but then $angle CFA=90$, which is a contradiction since the sum of the angles of a triangle is $leq180$. If $angle BCA<90$ then $angle FCAgt90$, but then $angle CFAgt90$, which is a contradiction for the same reason. In conclusion it must be $angle BCAgt90.$

share|cite|improve this answer

edited Jul 28 at 23:49

answered Jul 28 at 23:41

Pedro Fernandes

186

add a comment | 

up vote
0
down vote



accepted

I managed to solve this! I came up with this drawing when I saw @idk answer.
Since $triangle ACF$ is isosceles we have that $angle FCA =angle CFA$.
If $angle BCA=90$ then $angle FCA=90,$ but then $angle CFA=90$, which is a contradiction since the sum of the angles of a triangle is $leq180$. If $angle BCA<90$ then $angle FCAgt90$, but then $angle CFAgt90$, which is a contradiction for the same reason. In conclusion it must be $angle BCAgt90.$

share|cite|improve this answer

edited Jul 28 at 23:49

answered Jul 28 at 23:41

Pedro Fernandes

186

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

I managed to solve this! I came up with this drawing when I saw @idk answer.
Since $triangle ACF$ is isosceles we have that $angle FCA =angle CFA$.
If $angle BCA=90$ then $angle FCA=90,$ but then $angle CFA=90$, which is a contradiction since the sum of the angles of a triangle is $leq180$. If $angle BCA<90$ then $angle FCAgt90$, but then $angle CFAgt90$, which is a contradiction for the same reason. In conclusion it must be $angle BCAgt90.$

share|cite|improve this answer

edited Jul 28 at 23:49

answered Jul 28 at 23:41

Pedro Fernandes

186

I managed to solve this! I came up with this drawing when I saw @idk answer.
Since $triangle ACF$ is isosceles we have that $angle FCA =angle CFA$.
If $angle BCA=90$ then $angle FCA=90,$ but then $angle CFA=90$, which is a contradiction since the sum of the angles of a triangle is $leq180$. If $angle BCA<90$ then $angle FCAgt90$, but then $angle CFAgt90$, which is a contradiction for the same reason. In conclusion it must be $angle BCAgt90.$

share|cite|improve this answer

edited Jul 28 at 23:49

answered Jul 28 at 23:41

Pedro Fernandes

186

share|cite|improve this answer

share|cite|improve this answer

edited Jul 28 at 23:49

edited Jul 28 at 23:49

edited Jul 28 at 23:49

answered Jul 28 at 23:41

Pedro Fernandes

186

answered Jul 28 at 23:41

Pedro Fernandes

186

answered Jul 28 at 23:41

Pedro Fernandes

186

186

add a comment | 

add a comment |