Congruence of angles axiomIn neutral geometry, can a family of parallel lines leave holes in the plane?Stuck on geometry provingWithout using angle measure how do I prove two lines are parallel to the same line are parallel to each other?AAS Congruence Included Side?Prove congruence of pentagons with all right angles.Inscribing rhombus in a triangle's angle in only eight compass-and-straightedge stepsLet $x_1$ be the length $DE$ and $x_2$ is the length $BC$. Prove $x_1<x_2$.Visual proof of isosceles base-angle congruency?Neutral geometry: If one triangle has angle-sum $180^circ$, then all triangles have angle-sum $180^circ$

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Novel set in the future, children cannot change the class they are born into, one class is made uneducated by associating books with pain



Congruence of angles axiom


In neutral geometry, can a family of parallel lines leave holes in the plane?Stuck on geometry provingWithout using angle measure how do I prove two lines are parallel to the same line are parallel to each other?AAS Congruence Included Side?Prove congruence of pentagons with all right angles.Inscribing rhombus in a triangle's angle in only eight compass-and-straightedge stepsLet $x_1$ be the length $DE$ and $x_2$ is the length $BC$. Prove $x_1<x_2$.Visual proof of isosceles base-angle congruency?Neutral geometry: If one triangle has angle-sum $180^circ$, then all triangles have angle-sum $180^circ$






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;








3














$begingroup$


In my geometry book, the following statement is provided as axiom of congruence.



Let $ABC$ and $A'B'C'$ be two triangles. If $AB equiv A'B'$, $AC equiv A'C'$ and $ angle BAC equiv B'A'C'$ then $ angle ABC equiv A'B'C'$



I honestly have no idea where this comes from. Note that we use this axiom to prove $SSS$, $ASA$ and $SAS$.



Could someone provide intuition for something like this? It has to be obvious if it's used as an axiom.. right? I'm having trouble memorising something like that and using it to prove things. Not to mention the order here is extremely important.



EDIT:
As an extra note in the book it says, if we change it to $AC equiv A'C'$ , $ AB equiv A'B'$ and $angle CAB = angle C'A'B'$ then $angle ACB equiv angle A'C'B'$










share|cite|improve this question











$endgroup$















  • $begingroup$
    Try to construct two non-congruent triangles with the given quantities—you won’t succeed.
    $endgroup$
    – Michael Hoppe
    8 hours ago

















3














$begingroup$


In my geometry book, the following statement is provided as axiom of congruence.



Let $ABC$ and $A'B'C'$ be two triangles. If $AB equiv A'B'$, $AC equiv A'C'$ and $ angle BAC equiv B'A'C'$ then $ angle ABC equiv A'B'C'$



I honestly have no idea where this comes from. Note that we use this axiom to prove $SSS$, $ASA$ and $SAS$.



Could someone provide intuition for something like this? It has to be obvious if it's used as an axiom.. right? I'm having trouble memorising something like that and using it to prove things. Not to mention the order here is extremely important.



EDIT:
As an extra note in the book it says, if we change it to $AC equiv A'C'$ , $ AB equiv A'B'$ and $angle CAB = angle C'A'B'$ then $angle ACB equiv angle A'C'B'$










share|cite|improve this question











$endgroup$















  • $begingroup$
    Try to construct two non-congruent triangles with the given quantities—you won’t succeed.
    $endgroup$
    – Michael Hoppe
    8 hours ago













3












3








3





$begingroup$


In my geometry book, the following statement is provided as axiom of congruence.



Let $ABC$ and $A'B'C'$ be two triangles. If $AB equiv A'B'$, $AC equiv A'C'$ and $ angle BAC equiv B'A'C'$ then $ angle ABC equiv A'B'C'$



I honestly have no idea where this comes from. Note that we use this axiom to prove $SSS$, $ASA$ and $SAS$.



Could someone provide intuition for something like this? It has to be obvious if it's used as an axiom.. right? I'm having trouble memorising something like that and using it to prove things. Not to mention the order here is extremely important.



EDIT:
As an extra note in the book it says, if we change it to $AC equiv A'C'$ , $ AB equiv A'B'$ and $angle CAB = angle C'A'B'$ then $angle ACB equiv angle A'C'B'$










share|cite|improve this question











$endgroup$




In my geometry book, the following statement is provided as axiom of congruence.



Let $ABC$ and $A'B'C'$ be two triangles. If $AB equiv A'B'$, $AC equiv A'C'$ and $ angle BAC equiv B'A'C'$ then $ angle ABC equiv A'B'C'$



I honestly have no idea where this comes from. Note that we use this axiom to prove $SSS$, $ASA$ and $SAS$.



Could someone provide intuition for something like this? It has to be obvious if it's used as an axiom.. right? I'm having trouble memorising something like that and using it to prove things. Not to mention the order here is extremely important.



EDIT:
As an extra note in the book it says, if we change it to $AC equiv A'C'$ , $ AB equiv A'B'$ and $angle CAB = angle C'A'B'$ then $angle ACB equiv angle A'C'B'$







geometry axioms congruences-geometry






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share|cite|improve this question








edited 7 hours ago







DreaDk

















asked 8 hours ago









DreaDkDreaDk

7081 gold badge3 silver badges18 bronze badges




7081 gold badge3 silver badges18 bronze badges














  • $begingroup$
    Try to construct two non-congruent triangles with the given quantities—you won’t succeed.
    $endgroup$
    – Michael Hoppe
    8 hours ago
















  • $begingroup$
    Try to construct two non-congruent triangles with the given quantities—you won’t succeed.
    $endgroup$
    – Michael Hoppe
    8 hours ago















$begingroup$
Try to construct two non-congruent triangles with the given quantities—you won’t succeed.
$endgroup$
– Michael Hoppe
8 hours ago




$begingroup$
Try to construct two non-congruent triangles with the given quantities—you won’t succeed.
$endgroup$
– Michael Hoppe
8 hours ago










3 Answers
3






active

oldest

votes


















1
















$begingroup$

Welcome to axiomatic mathematics!



Axioms are accepted not just on the basis of obviousness.



Yes, the truth of this statement should be obvious, as the comment of @MichaelHoppe remarks.



And yes, axioms and other mathematical statements can have a persnickety precision to them which requires you to memorize symbols in order. All I'll say to this is, you memorized your counting numbers once upon a time when you were much much younger, I'm sure you can memorize this axiom.



But the real point about formulating axioms is that they should be as narrowly formulated as possible, subject to the all important restriction that the entire rest of the theory can be derived from the axioms. Mathematicians HATE to accept things without proof. They just HATE it. If they have to accept an axiom, they want it to be formulated as narrowly as possible. In the case of this axiom, the hypotheses are the same as the stronger SAS theorem, but the conclusion is quite a bit narrower: all the conclusion asserts is equality of one specific angle pair.



The point is that from this narrow conclusion, many broader conclusions may be derived. For example, from the narrow conclusion of equality of one angle pair, one may prove equality of the other angle pair, as you have correctly noted in the edit at the bottom of your question. And then one can go on to prove other theorems about triangles such as SSS and so on. All from just one, narrow, very specifically formulated, special version of SAS! A mathematician loves that.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    I have no doubt about the validity of the axiom. But I actually can't make sense of the order of the angles. The first angle is obviously the one between the sides, but the latter one appears random to me.
    $endgroup$
    – DreaDk
    6 hours ago






  • 1




    $begingroup$
    Well, the triangle only has three angles. Pick one of those two, and formulate your axiom with that one. As I've tried to say, the point is not the validity of the axiom, it is instead the narrowness of its conclusions, and the broadness of its applications.
    $endgroup$
    – Lee Mosher
    5 hours ago


















3
















$begingroup$

I am not sure this is a satisfactory answer.



That assertion is essentially Euclid's Proposition 4 (SAS) in Book I:




If two triangles have two sides equal to two sides respectively, and
have the angles contained by the equal straight lines equal, then they
also have the base equal to the base, the triangle equals the
triangle, and the remaining angles equal the remaining angles
respectively, namely those opposite the equal sides.




The notes on Euclid's proof at
https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI4.html comment clearly on his use of an essentially undefined notion of congruence based on superimposing one triangle on another.



I suspect the author of your text was sufficiently unhappy with Euclid's argument that they created a new axiom instead, in order to get started on the interesting theorems.



You will probably never need to use it directly as an axiom once you've used it to prove SSS, ASA and SAS.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    Still can't figure out the logic behind the equality of the angles ABC and A'B'C'. Do I just randomly pick one of the two remaining angles and say they are equal?
    $endgroup$
    – DreaDk
    7 hours ago










  • $begingroup$
    @DreaDk It seems as if that's what the book's author wants you to accept as an axiom. I don't like it either, but I understand why something like it is necessary in order to get to interesting geometry. Foundations are hard. See Hilbert's axioms at en.wikipedia.org/wiki/Hilbert%27s_axioms
    $endgroup$
    – Ethan Bolker
    7 hours ago


















0
















$begingroup$

The assumptions are identical to $SAS$ so why is it an Axiom ? They should be able to prove it from previous theorems.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    It's a weaker version of SAS that we use to prove SAS. This whole thing sounds so arbitrary when I read it.
    $endgroup$
    – DreaDk
    7 hours ago












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1
















$begingroup$

Welcome to axiomatic mathematics!



Axioms are accepted not just on the basis of obviousness.



Yes, the truth of this statement should be obvious, as the comment of @MichaelHoppe remarks.



And yes, axioms and other mathematical statements can have a persnickety precision to them which requires you to memorize symbols in order. All I'll say to this is, you memorized your counting numbers once upon a time when you were much much younger, I'm sure you can memorize this axiom.



But the real point about formulating axioms is that they should be as narrowly formulated as possible, subject to the all important restriction that the entire rest of the theory can be derived from the axioms. Mathematicians HATE to accept things without proof. They just HATE it. If they have to accept an axiom, they want it to be formulated as narrowly as possible. In the case of this axiom, the hypotheses are the same as the stronger SAS theorem, but the conclusion is quite a bit narrower: all the conclusion asserts is equality of one specific angle pair.



The point is that from this narrow conclusion, many broader conclusions may be derived. For example, from the narrow conclusion of equality of one angle pair, one may prove equality of the other angle pair, as you have correctly noted in the edit at the bottom of your question. And then one can go on to prove other theorems about triangles such as SSS and so on. All from just one, narrow, very specifically formulated, special version of SAS! A mathematician loves that.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    I have no doubt about the validity of the axiom. But I actually can't make sense of the order of the angles. The first angle is obviously the one between the sides, but the latter one appears random to me.
    $endgroup$
    – DreaDk
    6 hours ago






  • 1




    $begingroup$
    Well, the triangle only has three angles. Pick one of those two, and formulate your axiom with that one. As I've tried to say, the point is not the validity of the axiom, it is instead the narrowness of its conclusions, and the broadness of its applications.
    $endgroup$
    – Lee Mosher
    5 hours ago















1
















$begingroup$

Welcome to axiomatic mathematics!



Axioms are accepted not just on the basis of obviousness.



Yes, the truth of this statement should be obvious, as the comment of @MichaelHoppe remarks.



And yes, axioms and other mathematical statements can have a persnickety precision to them which requires you to memorize symbols in order. All I'll say to this is, you memorized your counting numbers once upon a time when you were much much younger, I'm sure you can memorize this axiom.



But the real point about formulating axioms is that they should be as narrowly formulated as possible, subject to the all important restriction that the entire rest of the theory can be derived from the axioms. Mathematicians HATE to accept things without proof. They just HATE it. If they have to accept an axiom, they want it to be formulated as narrowly as possible. In the case of this axiom, the hypotheses are the same as the stronger SAS theorem, but the conclusion is quite a bit narrower: all the conclusion asserts is equality of one specific angle pair.



The point is that from this narrow conclusion, many broader conclusions may be derived. For example, from the narrow conclusion of equality of one angle pair, one may prove equality of the other angle pair, as you have correctly noted in the edit at the bottom of your question. And then one can go on to prove other theorems about triangles such as SSS and so on. All from just one, narrow, very specifically formulated, special version of SAS! A mathematician loves that.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    I have no doubt about the validity of the axiom. But I actually can't make sense of the order of the angles. The first angle is obviously the one between the sides, but the latter one appears random to me.
    $endgroup$
    – DreaDk
    6 hours ago






  • 1




    $begingroup$
    Well, the triangle only has three angles. Pick one of those two, and formulate your axiom with that one. As I've tried to say, the point is not the validity of the axiom, it is instead the narrowness of its conclusions, and the broadness of its applications.
    $endgroup$
    – Lee Mosher
    5 hours ago













1














1










1







$begingroup$

Welcome to axiomatic mathematics!



Axioms are accepted not just on the basis of obviousness.



Yes, the truth of this statement should be obvious, as the comment of @MichaelHoppe remarks.



And yes, axioms and other mathematical statements can have a persnickety precision to them which requires you to memorize symbols in order. All I'll say to this is, you memorized your counting numbers once upon a time when you were much much younger, I'm sure you can memorize this axiom.



But the real point about formulating axioms is that they should be as narrowly formulated as possible, subject to the all important restriction that the entire rest of the theory can be derived from the axioms. Mathematicians HATE to accept things without proof. They just HATE it. If they have to accept an axiom, they want it to be formulated as narrowly as possible. In the case of this axiom, the hypotheses are the same as the stronger SAS theorem, but the conclusion is quite a bit narrower: all the conclusion asserts is equality of one specific angle pair.



The point is that from this narrow conclusion, many broader conclusions may be derived. For example, from the narrow conclusion of equality of one angle pair, one may prove equality of the other angle pair, as you have correctly noted in the edit at the bottom of your question. And then one can go on to prove other theorems about triangles such as SSS and so on. All from just one, narrow, very specifically formulated, special version of SAS! A mathematician loves that.






share|cite|improve this answer










$endgroup$



Welcome to axiomatic mathematics!



Axioms are accepted not just on the basis of obviousness.



Yes, the truth of this statement should be obvious, as the comment of @MichaelHoppe remarks.



And yes, axioms and other mathematical statements can have a persnickety precision to them which requires you to memorize symbols in order. All I'll say to this is, you memorized your counting numbers once upon a time when you were much much younger, I'm sure you can memorize this axiom.



But the real point about formulating axioms is that they should be as narrowly formulated as possible, subject to the all important restriction that the entire rest of the theory can be derived from the axioms. Mathematicians HATE to accept things without proof. They just HATE it. If they have to accept an axiom, they want it to be formulated as narrowly as possible. In the case of this axiom, the hypotheses are the same as the stronger SAS theorem, but the conclusion is quite a bit narrower: all the conclusion asserts is equality of one specific angle pair.



The point is that from this narrow conclusion, many broader conclusions may be derived. For example, from the narrow conclusion of equality of one angle pair, one may prove equality of the other angle pair, as you have correctly noted in the edit at the bottom of your question. And then one can go on to prove other theorems about triangles such as SSS and so on. All from just one, narrow, very specifically formulated, special version of SAS! A mathematician loves that.







share|cite|improve this answer













share|cite|improve this answer




share|cite|improve this answer



share|cite|improve this answer










answered 7 hours ago









Lee MosherLee Mosher

60.4k4 gold badges40 silver badges97 bronze badges




60.4k4 gold badges40 silver badges97 bronze badges














  • $begingroup$
    I have no doubt about the validity of the axiom. But I actually can't make sense of the order of the angles. The first angle is obviously the one between the sides, but the latter one appears random to me.
    $endgroup$
    – DreaDk
    6 hours ago






  • 1




    $begingroup$
    Well, the triangle only has three angles. Pick one of those two, and formulate your axiom with that one. As I've tried to say, the point is not the validity of the axiom, it is instead the narrowness of its conclusions, and the broadness of its applications.
    $endgroup$
    – Lee Mosher
    5 hours ago
















  • $begingroup$
    I have no doubt about the validity of the axiom. But I actually can't make sense of the order of the angles. The first angle is obviously the one between the sides, but the latter one appears random to me.
    $endgroup$
    – DreaDk
    6 hours ago






  • 1




    $begingroup$
    Well, the triangle only has three angles. Pick one of those two, and formulate your axiom with that one. As I've tried to say, the point is not the validity of the axiom, it is instead the narrowness of its conclusions, and the broadness of its applications.
    $endgroup$
    – Lee Mosher
    5 hours ago















$begingroup$
I have no doubt about the validity of the axiom. But I actually can't make sense of the order of the angles. The first angle is obviously the one between the sides, but the latter one appears random to me.
$endgroup$
– DreaDk
6 hours ago




$begingroup$
I have no doubt about the validity of the axiom. But I actually can't make sense of the order of the angles. The first angle is obviously the one between the sides, but the latter one appears random to me.
$endgroup$
– DreaDk
6 hours ago




1




1




$begingroup$
Well, the triangle only has three angles. Pick one of those two, and formulate your axiom with that one. As I've tried to say, the point is not the validity of the axiom, it is instead the narrowness of its conclusions, and the broadness of its applications.
$endgroup$
– Lee Mosher
5 hours ago




$begingroup$
Well, the triangle only has three angles. Pick one of those two, and formulate your axiom with that one. As I've tried to say, the point is not the validity of the axiom, it is instead the narrowness of its conclusions, and the broadness of its applications.
$endgroup$
– Lee Mosher
5 hours ago













3
















$begingroup$

I am not sure this is a satisfactory answer.



That assertion is essentially Euclid's Proposition 4 (SAS) in Book I:




If two triangles have two sides equal to two sides respectively, and
have the angles contained by the equal straight lines equal, then they
also have the base equal to the base, the triangle equals the
triangle, and the remaining angles equal the remaining angles
respectively, namely those opposite the equal sides.




The notes on Euclid's proof at
https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI4.html comment clearly on his use of an essentially undefined notion of congruence based on superimposing one triangle on another.



I suspect the author of your text was sufficiently unhappy with Euclid's argument that they created a new axiom instead, in order to get started on the interesting theorems.



You will probably never need to use it directly as an axiom once you've used it to prove SSS, ASA and SAS.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    Still can't figure out the logic behind the equality of the angles ABC and A'B'C'. Do I just randomly pick one of the two remaining angles and say they are equal?
    $endgroup$
    – DreaDk
    7 hours ago










  • $begingroup$
    @DreaDk It seems as if that's what the book's author wants you to accept as an axiom. I don't like it either, but I understand why something like it is necessary in order to get to interesting geometry. Foundations are hard. See Hilbert's axioms at en.wikipedia.org/wiki/Hilbert%27s_axioms
    $endgroup$
    – Ethan Bolker
    7 hours ago















3
















$begingroup$

I am not sure this is a satisfactory answer.



That assertion is essentially Euclid's Proposition 4 (SAS) in Book I:




If two triangles have two sides equal to two sides respectively, and
have the angles contained by the equal straight lines equal, then they
also have the base equal to the base, the triangle equals the
triangle, and the remaining angles equal the remaining angles
respectively, namely those opposite the equal sides.




The notes on Euclid's proof at
https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI4.html comment clearly on his use of an essentially undefined notion of congruence based on superimposing one triangle on another.



I suspect the author of your text was sufficiently unhappy with Euclid's argument that they created a new axiom instead, in order to get started on the interesting theorems.



You will probably never need to use it directly as an axiom once you've used it to prove SSS, ASA and SAS.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    Still can't figure out the logic behind the equality of the angles ABC and A'B'C'. Do I just randomly pick one of the two remaining angles and say they are equal?
    $endgroup$
    – DreaDk
    7 hours ago










  • $begingroup$
    @DreaDk It seems as if that's what the book's author wants you to accept as an axiom. I don't like it either, but I understand why something like it is necessary in order to get to interesting geometry. Foundations are hard. See Hilbert's axioms at en.wikipedia.org/wiki/Hilbert%27s_axioms
    $endgroup$
    – Ethan Bolker
    7 hours ago













3














3










3







$begingroup$

I am not sure this is a satisfactory answer.



That assertion is essentially Euclid's Proposition 4 (SAS) in Book I:




If two triangles have two sides equal to two sides respectively, and
have the angles contained by the equal straight lines equal, then they
also have the base equal to the base, the triangle equals the
triangle, and the remaining angles equal the remaining angles
respectively, namely those opposite the equal sides.




The notes on Euclid's proof at
https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI4.html comment clearly on his use of an essentially undefined notion of congruence based on superimposing one triangle on another.



I suspect the author of your text was sufficiently unhappy with Euclid's argument that they created a new axiom instead, in order to get started on the interesting theorems.



You will probably never need to use it directly as an axiom once you've used it to prove SSS, ASA and SAS.






share|cite|improve this answer










$endgroup$



I am not sure this is a satisfactory answer.



That assertion is essentially Euclid's Proposition 4 (SAS) in Book I:




If two triangles have two sides equal to two sides respectively, and
have the angles contained by the equal straight lines equal, then they
also have the base equal to the base, the triangle equals the
triangle, and the remaining angles equal the remaining angles
respectively, namely those opposite the equal sides.




The notes on Euclid's proof at
https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI4.html comment clearly on his use of an essentially undefined notion of congruence based on superimposing one triangle on another.



I suspect the author of your text was sufficiently unhappy with Euclid's argument that they created a new axiom instead, in order to get started on the interesting theorems.



You will probably never need to use it directly as an axiom once you've used it to prove SSS, ASA and SAS.







share|cite|improve this answer













share|cite|improve this answer




share|cite|improve this answer



share|cite|improve this answer










answered 8 hours ago









Ethan BolkerEthan Bolker

56.8k5 gold badges63 silver badges138 bronze badges




56.8k5 gold badges63 silver badges138 bronze badges














  • $begingroup$
    Still can't figure out the logic behind the equality of the angles ABC and A'B'C'. Do I just randomly pick one of the two remaining angles and say they are equal?
    $endgroup$
    – DreaDk
    7 hours ago










  • $begingroup$
    @DreaDk It seems as if that's what the book's author wants you to accept as an axiom. I don't like it either, but I understand why something like it is necessary in order to get to interesting geometry. Foundations are hard. See Hilbert's axioms at en.wikipedia.org/wiki/Hilbert%27s_axioms
    $endgroup$
    – Ethan Bolker
    7 hours ago
















  • $begingroup$
    Still can't figure out the logic behind the equality of the angles ABC and A'B'C'. Do I just randomly pick one of the two remaining angles and say they are equal?
    $endgroup$
    – DreaDk
    7 hours ago










  • $begingroup$
    @DreaDk It seems as if that's what the book's author wants you to accept as an axiom. I don't like it either, but I understand why something like it is necessary in order to get to interesting geometry. Foundations are hard. See Hilbert's axioms at en.wikipedia.org/wiki/Hilbert%27s_axioms
    $endgroup$
    – Ethan Bolker
    7 hours ago















$begingroup$
Still can't figure out the logic behind the equality of the angles ABC and A'B'C'. Do I just randomly pick one of the two remaining angles and say they are equal?
$endgroup$
– DreaDk
7 hours ago




$begingroup$
Still can't figure out the logic behind the equality of the angles ABC and A'B'C'. Do I just randomly pick one of the two remaining angles and say they are equal?
$endgroup$
– DreaDk
7 hours ago












$begingroup$
@DreaDk It seems as if that's what the book's author wants you to accept as an axiom. I don't like it either, but I understand why something like it is necessary in order to get to interesting geometry. Foundations are hard. See Hilbert's axioms at en.wikipedia.org/wiki/Hilbert%27s_axioms
$endgroup$
– Ethan Bolker
7 hours ago




$begingroup$
@DreaDk It seems as if that's what the book's author wants you to accept as an axiom. I don't like it either, but I understand why something like it is necessary in order to get to interesting geometry. Foundations are hard. See Hilbert's axioms at en.wikipedia.org/wiki/Hilbert%27s_axioms
$endgroup$
– Ethan Bolker
7 hours ago











0
















$begingroup$

The assumptions are identical to $SAS$ so why is it an Axiom ? They should be able to prove it from previous theorems.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    It's a weaker version of SAS that we use to prove SAS. This whole thing sounds so arbitrary when I read it.
    $endgroup$
    – DreaDk
    7 hours ago















0
















$begingroup$

The assumptions are identical to $SAS$ so why is it an Axiom ? They should be able to prove it from previous theorems.






share|cite|improve this answer










$endgroup$














  • $begingroup$
    It's a weaker version of SAS that we use to prove SAS. This whole thing sounds so arbitrary when I read it.
    $endgroup$
    – DreaDk
    7 hours ago













0














0










0







$begingroup$

The assumptions are identical to $SAS$ so why is it an Axiom ? They should be able to prove it from previous theorems.






share|cite|improve this answer










$endgroup$



The assumptions are identical to $SAS$ so why is it an Axiom ? They should be able to prove it from previous theorems.







share|cite|improve this answer













share|cite|improve this answer




share|cite|improve this answer



share|cite|improve this answer










answered 7 hours ago









Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.4k4 gold badges27 silver badges74 bronze badges




54.4k4 gold badges27 silver badges74 bronze badges














  • $begingroup$
    It's a weaker version of SAS that we use to prove SAS. This whole thing sounds so arbitrary when I read it.
    $endgroup$
    – DreaDk
    7 hours ago
















  • $begingroup$
    It's a weaker version of SAS that we use to prove SAS. This whole thing sounds so arbitrary when I read it.
    $endgroup$
    – DreaDk
    7 hours ago















$begingroup$
It's a weaker version of SAS that we use to prove SAS. This whole thing sounds so arbitrary when I read it.
$endgroup$
– DreaDk
7 hours ago




$begingroup$
It's a weaker version of SAS that we use to prove SAS. This whole thing sounds so arbitrary when I read it.
$endgroup$
– DreaDk
7 hours ago


















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