Mixing
Why does having alternate interior angles congruent, etc., prove that two lines are parallel?
$begingroup$
I know that if two lines are parallel and there is a transversal crossing both, the alternate interior angles are congruent, alternate exterior angles congruent, etc. etc.
According to the geometry textbook that the student I'm tutoring brought, the converse is true as well: if the alternate interior angles are congruent, or the consecutive interior angles are supplementary, etc., then the lines must be parallel.
Why should this be? I mean, it looks like it should be true, but what is the proof that it's true?
euclidean-geometry
asked Jan 9 at 2:25
DonielFDonielF
484515
$endgroup$
add a comment |
$begingroup$
I know that if two lines are parallel and there is a transversal crossing both, the alternate interior angles are congruent, alternate exterior angles congruent, etc. etc.
According to the geometry textbook that the student I'm tutoring brought, the converse is true as well: if the alternate interior angles are congruent, or the consecutive interior angles are supplementary, etc., then the lines must be parallel.
Why should this be? I mean, it looks like it should be true, but what is the proof that it's true?
euclidean-geometry
asked Jan 9 at 2:25
DonielFDonielF
484515
$endgroup$
add a comment |
$begingroup$
I know that if two lines are parallel and there is a transversal crossing both, the alternate interior angles are congruent, alternate exterior angles congruent, etc. etc.
According to the geometry textbook that the student I'm tutoring brought, the converse is true as well: if the alternate interior angles are congruent, or the consecutive interior angles are supplementary, etc., then the lines must be parallel.
Why should this be? I mean, it looks like it should be true, but what is the proof that it's true?
euclidean-geometry
asked Jan 9 at 2:25
DonielFDonielF
484515
$endgroup$
I know that if two lines are parallel and there is a transversal crossing both, the alternate interior angles are congruent, alternate exterior angles congruent, etc. etc.
According to the geometry textbook that the student I'm tutoring brought, the converse is true as well: if the alternate interior angles are congruent, or the consecutive interior angles are supplementary, etc., then the lines must be parallel.
Why should this be? I mean, it looks like it should be true, but what is the proof that it's true?
euclidean-geometry
euclidean-geometry
asked Jan 9 at 2:25
DonielFDonielF
484515
asked Jan 9 at 2:25
DonielFDonielF
484515
asked Jan 9 at 2:25
DonielFDonielF
484515
asked Jan 9 at 2:25
DonielFDonielF
484515
asked Jan 9 at 2:25
DonielFDonielF
484515
484515
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Assuming you are working from some variation of Euclid's axioms, you can prove this as follows. Suppose two lines $ell$ and $m$ are not parallel but form congruent alternate interior angles with a third line $n$. Since $ell$ and $m$ are not parallel, we can thus form a triangle whose sides are segments of $ell$, $m$, and $n$. But by hypothesis, the two angles of this triangle along the segment of $n$ are supplementary. This is a contradiction, since two angles in a triangle must add up to less than $pi$.
(Note that some care is needed to avoid circularity in this last step: the proofs that I know of for the fact that the sum of the angles in a triangle is $pi$ use the existence of parallel lines, and the classical construction of parallel lines uses the statement we are trying to prove here. However, the statement that two angles in a triangle add up to less than $pi$ can be proved without using parallel lines; see for instance Euclid Book I, Proposition 17 which is a corollary of Proposition 16. Alternatively, in axiomatizations of geometry that include a Dedekind-style completeness axiom, the existence of parallel lines can be proved from completeness.)
answered Jan 9 at 3:41
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
Is the parenthetical statement true? If we start with parallel lines yields alternate interior etc., and use that to prove that triangles add to π, why can’t we use that to prove that alternate interior prove parallel lines as you describe? You’re not using the statement to prove itself, you’re using the statement to prove its converse.
$endgroup$
– DonielF
Jan 9 at 3:46
$begingroup$
The proof that triangles add to $pi$ uses both the statement and its converse. Essentially, it uses the fact that if you have a pair of lines which have equal alternate interior angles for one transversal, they also have equal alternate interior angles for any other transversal. So you use the first transversal (and the statement in one direction) to conclude the lines are parallel, and then the statement in the other direction to get the conclusion for the other transversal.
$endgroup$
– Eric Wofsey
Jan 9 at 3:48
$begingroup$
I’ve never seen that version before. The classic proof I’ve heard simply constructs a line at one angle of the triangle such that it is parallel to the third side, and proof is by alternate interior. No need to prove that the lines are parallel, as it’s constructed originally such that it is.
$endgroup$
– DonielF
Jan 9 at 3:51
$begingroup$
But how do you know the construction is correct? That is, that the line constructed really is parallel? You need to have some way of proving that two lines are parallel.
$endgroup$
– Eric Wofsey
Jan 9 at 3:53
add a comment |
$begingroup$
The statement "if consecutive interior angles are supplementary then lines are parallel" follows directly from the Euclid's fifth postulate which says that if the sum of two interior angles is less than two right angles, the lines will intersect. Using this postulate and considering straight angles formed by the transverse you can prove that if alternate interior angles are congruent, the lines are parallel. Below is the link to Euclid's Elements, I think the postulate is in Book 1.
http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf
Below is the proof for congruent alternate interior angles:
answered Jan 9 at 2:57
VasyaVasya
3,3201515
$endgroup$
$begingroup$
This is wrong: the fifth postulate is the converse of the desired statement. The desired statement is Book 1 Proposition 27, whose proof does not use the fifth postulate.
$endgroup$
– Eric Wofsey
Jan 9 at 3:07
$begingroup$
@EricWofsey: I think you need to be more specific what is wrong. I wasn't talking about the proposition 27, only the fifth postulate. I can prove statement regarding alternate interior angles using it.
$endgroup$
– Vasya
Jan 9 at 3:19
$begingroup$
How would you prove it? The proof certainly is not obvious as far as I can tell.
$endgroup$
– Eric Wofsey
Jan 9 at 3:20
$begingroup$
if alternate interior angles are congruent then alternate exterior angles are congruent (as they are complementary to the interior). Then using straight angles and transitivity we can show that the same side interior/exterior angles are congruent.
$endgroup$
– Vasya
Jan 9 at 3:41
$begingroup$
And how do you conclude that the lines are parallel?
$endgroup$
– Eric Wofsey
Jan 9 at 3:51
|
show 3 more comments
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2 Answers
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2 Answers
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$begingroup$
Assuming you are working from some variation of Euclid's axioms, you can prove this as follows. Suppose two lines $ell$ and $m$ are not parallel but form congruent alternate interior angles with a third line $n$. Since $ell$ and $m$ are not parallel, we can thus form a triangle whose sides are segments of $ell$, $m$, and $n$. But by hypothesis, the two angles of this triangle along the segment of $n$ are supplementary. This is a contradiction, since two angles in a triangle must add up to less than $pi$.
(Note that some care is needed to avoid circularity in this last step: the proofs that I know of for the fact that the sum of the angles in a triangle is $pi$ use the existence of parallel lines, and the classical construction of parallel lines uses the statement we are trying to prove here. However, the statement that two angles in a triangle add up to less than $pi$ can be proved without using parallel lines; see for instance Euclid Book I, Proposition 17 which is a corollary of Proposition 16. Alternatively, in axiomatizations of geometry that include a Dedekind-style completeness axiom, the existence of parallel lines can be proved from completeness.)
answered Jan 9 at 3:41
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
Is the parenthetical statement true? If we start with parallel lines yields alternate interior etc., and use that to prove that triangles add to π, why can’t we use that to prove that alternate interior prove parallel lines as you describe? You’re not using the statement to prove itself, you’re using the statement to prove its converse.
$endgroup$
– DonielF
Jan 9 at 3:46
$begingroup$
The proof that triangles add to $pi$ uses both the statement and its converse. Essentially, it uses the fact that if you have a pair of lines which have equal alternate interior angles for one transversal, they also have equal alternate interior angles for any other transversal. So you use the first transversal (and the statement in one direction) to conclude the lines are parallel, and then the statement in the other direction to get the conclusion for the other transversal.
$endgroup$
– Eric Wofsey
Jan 9 at 3:48
$begingroup$
I’ve never seen that version before. The classic proof I’ve heard simply constructs a line at one angle of the triangle such that it is parallel to the third side, and proof is by alternate interior. No need to prove that the lines are parallel, as it’s constructed originally such that it is.
$endgroup$
– DonielF
Jan 9 at 3:51
$begingroup$
But how do you know the construction is correct? That is, that the line constructed really is parallel? You need to have some way of proving that two lines are parallel.
$endgroup$
– Eric Wofsey
Jan 9 at 3:53
add a comment |
$begingroup$
Assuming you are working from some variation of Euclid's axioms, you can prove this as follows. Suppose two lines $ell$ and $m$ are not parallel but form congruent alternate interior angles with a third line $n$. Since $ell$ and $m$ are not parallel, we can thus form a triangle whose sides are segments of $ell$, $m$, and $n$. But by hypothesis, the two angles of this triangle along the segment of $n$ are supplementary. This is a contradiction, since two angles in a triangle must add up to less than $pi$.
(Note that some care is needed to avoid circularity in this last step: the proofs that I know of for the fact that the sum of the angles in a triangle is $pi$ use the existence of parallel lines, and the classical construction of parallel lines uses the statement we are trying to prove here. However, the statement that two angles in a triangle add up to less than $pi$ can be proved without using parallel lines; see for instance Euclid Book I, Proposition 17 which is a corollary of Proposition 16. Alternatively, in axiomatizations of geometry that include a Dedekind-style completeness axiom, the existence of parallel lines can be proved from completeness.)
answered Jan 9 at 3:41
Eric WofseyEric Wofsey
184k13212338
$endgroup$
$begingroup$
Is the parenthetical statement true? If we start with parallel lines yields alternate interior etc., and use that to prove that triangles add to π, why can’t we use that to prove that alternate interior prove parallel lines as you describe? You’re not using the statement to prove itself, you’re using the statement to prove its converse.
$endgroup$
– DonielF
Jan 9 at 3:46
$begingroup$
The proof that triangles add to $pi$ uses both the statement and its converse. Essentially, it uses the fact that if you have a pair of lines which have equal alternate interior angles for one transversal, they also have equal alternate interior angles for any other transversal. So you use the first transversal (and the statement in one direction) to conclude the lines are parallel, and then the statement in the other direction to get the conclusion for the other transversal.
$endgroup$
– Eric Wofsey
Jan 9 at 3:48
$begingroup$
I’ve never seen that version before. The classic proof I’ve heard simply constructs a line at one angle of the triangle such that it is parallel to the third side, and proof is by alternate interior. No need to prove that the lines are parallel, as it’s constructed originally such that it is.
$endgroup$
– DonielF
Jan 9 at 3:51
$begingroup$
But how do you know the construction is correct? That is, that the line constructed really is parallel? You need to have some way of proving that two lines are parallel.
$endgroup$
– Eric Wofsey
Jan 9 at 3:53
add a comment |
$begingroup$
Assuming you are working from some variation of Euclid's axioms, you can prove this as follows. Suppose two lines $ell$ and $m$ are not parallel but form congruent alternate interior angles with a third line $n$. Since $ell$ and $m$ are not parallel, we can thus form a triangle whose sides are segments of $ell$, $m$, and $n$. But by hypothesis, the two angles of this triangle along the segment of $n$ are supplementary. This is a contradiction, since two angles in a triangle must add up to less than $pi$.
(Note that some care is needed to avoid circularity in this last step: the proofs that I know of for the fact that the sum of the angles in a triangle is $pi$ use the existence of parallel lines, and the classical construction of parallel lines uses the statement we are trying to prove here. However, the statement that two angles in a triangle add up to less than $pi$ can be proved without using parallel lines; see for instance Euclid Book I, Proposition 17 which is a corollary of Proposition 16. Alternatively, in axiomatizations of geometry that include a Dedekind-style completeness axiom, the existence of parallel lines can be proved from completeness.)
answered Jan 9 at 3:41
Eric WofseyEric Wofsey
184k13212338
$endgroup$
Assuming you are working from some variation of Euclid's axioms, you can prove this as follows. Suppose two lines $ell$ and $m$ are not parallel but form congruent alternate interior angles with a third line $n$. Since $ell$ and $m$ are not parallel, we can thus form a triangle whose sides are segments of $ell$, $m$, and $n$. But by hypothesis, the two angles of this triangle along the segment of $n$ are supplementary. This is a contradiction, since two angles in a triangle must add up to less than $pi$.
(Note that some care is needed to avoid circularity in this last step: the proofs that I know of for the fact that the sum of the angles in a triangle is $pi$ use the existence of parallel lines, and the classical construction of parallel lines uses the statement we are trying to prove here. However, the statement that two angles in a triangle add up to less than $pi$ can be proved without using parallel lines; see for instance Euclid Book I, Proposition 17 which is a corollary of Proposition 16. Alternatively, in axiomatizations of geometry that include a Dedekind-style completeness axiom, the existence of parallel lines can be proved from completeness.)
answered Jan 9 at 3:41
Eric WofseyEric Wofsey
184k13212338
answered Jan 9 at 3:41
Eric WofseyEric Wofsey
184k13212338
answered Jan 9 at 3:41
Eric WofseyEric Wofsey
184k13212338
answered Jan 9 at 3:41
Eric WofseyEric Wofsey
184k13212338
184k13212338
$begingroup$
Is the parenthetical statement true? If we start with parallel lines yields alternate interior etc., and use that to prove that triangles add to π, why can’t we use that to prove that alternate interior prove parallel lines as you describe? You’re not using the statement to prove itself, you’re using the statement to prove its converse.
$endgroup$
– DonielF
Jan 9 at 3:46
$begingroup$
The proof that triangles add to $pi$ uses both the statement and its converse. Essentially, it uses the fact that if you have a pair of lines which have equal alternate interior angles for one transversal, they also have equal alternate interior angles for any other transversal. So you use the first transversal (and the statement in one direction) to conclude the lines are parallel, and then the statement in the other direction to get the conclusion for the other transversal.
$endgroup$
– Eric Wofsey
Jan 9 at 3:48
$begingroup$
I’ve never seen that version before. The classic proof I’ve heard simply constructs a line at one angle of the triangle such that it is parallel to the third side, and proof is by alternate interior. No need to prove that the lines are parallel, as it’s constructed originally such that it is.
$endgroup$
– DonielF
Jan 9 at 3:51
$begingroup$
But how do you know the construction is correct? That is, that the line constructed really is parallel? You need to have some way of proving that two lines are parallel.
$endgroup$
– Eric Wofsey
Jan 9 at 3:53
add a comment |
$begingroup$
Is the parenthetical statement true? If we start with parallel lines yields alternate interior etc., and use that to prove that triangles add to π, why can’t we use that to prove that alternate interior prove parallel lines as you describe? You’re not using the statement to prove itself, you’re using the statement to prove its converse.
$endgroup$
– DonielF
Jan 9 at 3:46
$begingroup$
The proof that triangles add to $pi$ uses both the statement and its converse. Essentially, it uses the fact that if you have a pair of lines which have equal alternate interior angles for one transversal, they also have equal alternate interior angles for any other transversal. So you use the first transversal (and the statement in one direction) to conclude the lines are parallel, and then the statement in the other direction to get the conclusion for the other transversal.
$endgroup$
– Eric Wofsey
Jan 9 at 3:48
$begingroup$
I’ve never seen that version before. The classic proof I’ve heard simply constructs a line at one angle of the triangle such that it is parallel to the third side, and proof is by alternate interior. No need to prove that the lines are parallel, as it’s constructed originally such that it is.
$endgroup$
– DonielF
Jan 9 at 3:51
$begingroup$
But how do you know the construction is correct? That is, that the line constructed really is parallel? You need to have some way of proving that two lines are parallel.
$endgroup$
– Eric Wofsey
Jan 9 at 3:53
$begingroup$
Is the parenthetical statement true? If we start with parallel lines yields alternate interior etc., and use that to prove that triangles add to π, why can’t we use that to prove that alternate interior prove parallel lines as you describe? You’re not using the statement to prove itself, you’re using the statement to prove its converse.
$endgroup$
– DonielF
Jan 9 at 3:46
$begingroup$
Is the parenthetical statement true? If we start with parallel lines yields alternate interior etc., and use that to prove that triangles add to π, why can’t we use that to prove that alternate interior prove parallel lines as you describe? You’re not using the statement to prove itself, you’re using the statement to prove its converse.
$endgroup$
– DonielF
Jan 9 at 3:46
$begingroup$
The proof that triangles add to $pi$ uses both the statement and its converse. Essentially, it uses the fact that if you have a pair of lines which have equal alternate interior angles for one transversal, they also have equal alternate interior angles for any other transversal. So you use the first transversal (and the statement in one direction) to conclude the lines are parallel, and then the statement in the other direction to get the conclusion for the other transversal.
$endgroup$
– Eric Wofsey
Jan 9 at 3:48
$begingroup$
The proof that triangles add to $pi$ uses both the statement and its converse. Essentially, it uses the fact that if you have a pair of lines which have equal alternate interior angles for one transversal, they also have equal alternate interior angles for any other transversal. So you use the first transversal (and the statement in one direction) to conclude the lines are parallel, and then the statement in the other direction to get the conclusion for the other transversal.
$endgroup$
– Eric Wofsey
Jan 9 at 3:48
$begingroup$
I’ve never seen that version before. The classic proof I’ve heard simply constructs a line at one angle of the triangle such that it is parallel to the third side, and proof is by alternate interior. No need to prove that the lines are parallel, as it’s constructed originally such that it is.
$endgroup$
– DonielF
Jan 9 at 3:51
$begingroup$
I’ve never seen that version before. The classic proof I’ve heard simply constructs a line at one angle of the triangle such that it is parallel to the third side, and proof is by alternate interior. No need to prove that the lines are parallel, as it’s constructed originally such that it is.
$endgroup$
– DonielF
Jan 9 at 3:51
$begingroup$
But how do you know the construction is correct? That is, that the line constructed really is parallel? You need to have some way of proving that two lines are parallel.
$endgroup$
– Eric Wofsey
Jan 9 at 3:53
$begingroup$
But how do you know the construction is correct? That is, that the line constructed really is parallel? You need to have some way of proving that two lines are parallel.
$endgroup$
– Eric Wofsey
Jan 9 at 3:53
add a comment |
$begingroup$
The statement "if consecutive interior angles are supplementary then lines are parallel" follows directly from the Euclid's fifth postulate which says that if the sum of two interior angles is less than two right angles, the lines will intersect. Using this postulate and considering straight angles formed by the transverse you can prove that if alternate interior angles are congruent, the lines are parallel. Below is the link to Euclid's Elements, I think the postulate is in Book 1.
http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf
Below is the proof for congruent alternate interior angles:
answered Jan 9 at 2:57
VasyaVasya
3,3201515
$endgroup$
$begingroup$
This is wrong: the fifth postulate is the converse of the desired statement. The desired statement is Book 1 Proposition 27, whose proof does not use the fifth postulate.
$endgroup$
– Eric Wofsey
Jan 9 at 3:07
$begingroup$
@EricWofsey: I think you need to be more specific what is wrong. I wasn't talking about the proposition 27, only the fifth postulate. I can prove statement regarding alternate interior angles using it.
$endgroup$
– Vasya
Jan 9 at 3:19
$begingroup$
How would you prove it? The proof certainly is not obvious as far as I can tell.
$endgroup$
– Eric Wofsey
Jan 9 at 3:20
$begingroup$
if alternate interior angles are congruent then alternate exterior angles are congruent (as they are complementary to the interior). Then using straight angles and transitivity we can show that the same side interior/exterior angles are congruent.
$endgroup$
– Vasya
Jan 9 at 3:41
$begingroup$
And how do you conclude that the lines are parallel?
$endgroup$
– Eric Wofsey
Jan 9 at 3:51
|
show 3 more comments
$begingroup$
The statement "if consecutive interior angles are supplementary then lines are parallel" follows directly from the Euclid's fifth postulate which says that if the sum of two interior angles is less than two right angles, the lines will intersect. Using this postulate and considering straight angles formed by the transverse you can prove that if alternate interior angles are congruent, the lines are parallel. Below is the link to Euclid's Elements, I think the postulate is in Book 1.
http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf
Below is the proof for congruent alternate interior angles:
answered Jan 9 at 2:57
VasyaVasya
3,3201515
$endgroup$
$begingroup$
This is wrong: the fifth postulate is the converse of the desired statement. The desired statement is Book 1 Proposition 27, whose proof does not use the fifth postulate.
$endgroup$
– Eric Wofsey
Jan 9 at 3:07
$begingroup$
@EricWofsey: I think you need to be more specific what is wrong. I wasn't talking about the proposition 27, only the fifth postulate. I can prove statement regarding alternate interior angles using it.
$endgroup$
– Vasya
Jan 9 at 3:19
$begingroup$
How would you prove it? The proof certainly is not obvious as far as I can tell.
$endgroup$
– Eric Wofsey
Jan 9 at 3:20
$begingroup$
if alternate interior angles are congruent then alternate exterior angles are congruent (as they are complementary to the interior). Then using straight angles and transitivity we can show that the same side interior/exterior angles are congruent.
$endgroup$
– Vasya
Jan 9 at 3:41
$begingroup$
And how do you conclude that the lines are parallel?
$endgroup$
– Eric Wofsey
Jan 9 at 3:51
|
show 3 more comments
$begingroup$
The statement "if consecutive interior angles are supplementary then lines are parallel" follows directly from the Euclid's fifth postulate which says that if the sum of two interior angles is less than two right angles, the lines will intersect. Using this postulate and considering straight angles formed by the transverse you can prove that if alternate interior angles are congruent, the lines are parallel. Below is the link to Euclid's Elements, I think the postulate is in Book 1.
http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf
Below is the proof for congruent alternate interior angles:
answered Jan 9 at 2:57
VasyaVasya
3,3201515
$endgroup$
The statement "if consecutive interior angles are supplementary then lines are parallel" follows directly from the Euclid's fifth postulate which says that if the sum of two interior angles is less than two right angles, the lines will intersect. Using this postulate and considering straight angles formed by the transverse you can prove that if alternate interior angles are congruent, the lines are parallel. Below is the link to Euclid's Elements, I think the postulate is in Book 1.
http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf
Below is the proof for congruent alternate interior angles:
answered Jan 9 at 2:57
VasyaVasya
3,3201515
edited Jan 10 at 3:45
answered Jan 9 at 2:57
VasyaVasya
3,3201515
answered Jan 9 at 2:57
VasyaVasya
3,3201515
answered Jan 9 at 2:57
VasyaVasya
3,3201515
3,3201515
$begingroup$
This is wrong: the fifth postulate is the converse of the desired statement. The desired statement is Book 1 Proposition 27, whose proof does not use the fifth postulate.
$endgroup$
– Eric Wofsey
Jan 9 at 3:07
$begingroup$
@EricWofsey: I think you need to be more specific what is wrong. I wasn't talking about the proposition 27, only the fifth postulate. I can prove statement regarding alternate interior angles using it.
$endgroup$
– Vasya
Jan 9 at 3:19
$begingroup$
How would you prove it? The proof certainly is not obvious as far as I can tell.
$endgroup$
– Eric Wofsey
Jan 9 at 3:20
$begingroup$
if alternate interior angles are congruent then alternate exterior angles are congruent (as they are complementary to the interior). Then using straight angles and transitivity we can show that the same side interior/exterior angles are congruent.
$endgroup$
– Vasya
Jan 9 at 3:41
$begingroup$
And how do you conclude that the lines are parallel?
$endgroup$
– Eric Wofsey
Jan 9 at 3:51
|
show 3 more comments
$begingroup$
This is wrong: the fifth postulate is the converse of the desired statement. The desired statement is Book 1 Proposition 27, whose proof does not use the fifth postulate.
$endgroup$
– Eric Wofsey
Jan 9 at 3:07
$begingroup$
@EricWofsey: I think you need to be more specific what is wrong. I wasn't talking about the proposition 27, only the fifth postulate. I can prove statement regarding alternate interior angles using it.
$endgroup$
– Vasya
Jan 9 at 3:19
$begingroup$
How would you prove it? The proof certainly is not obvious as far as I can tell.
$endgroup$
– Eric Wofsey
Jan 9 at 3:20
$begingroup$
if alternate interior angles are congruent then alternate exterior angles are congruent (as they are complementary to the interior). Then using straight angles and transitivity we can show that the same side interior/exterior angles are congruent.
$endgroup$
– Vasya
Jan 9 at 3:41
$begingroup$
And how do you conclude that the lines are parallel?
$endgroup$
– Eric Wofsey
Jan 9 at 3:51
$begingroup$
This is wrong: the fifth postulate is the converse of the desired statement. The desired statement is Book 1 Proposition 27, whose proof does not use the fifth postulate.
$endgroup$
– Eric Wofsey
Jan 9 at 3:07
$begingroup$
This is wrong: the fifth postulate is the converse of the desired statement. The desired statement is Book 1 Proposition 27, whose proof does not use the fifth postulate.
$endgroup$
– Eric Wofsey
Jan 9 at 3:07
$begingroup$
@EricWofsey: I think you need to be more specific what is wrong. I wasn't talking about the proposition 27, only the fifth postulate. I can prove statement regarding alternate interior angles using it.
$endgroup$
– Vasya
Jan 9 at 3:19
$begingroup$
@EricWofsey: I think you need to be more specific what is wrong. I wasn't talking about the proposition 27, only the fifth postulate. I can prove statement regarding alternate interior angles using it.
$endgroup$
– Vasya
Jan 9 at 3:19
$begingroup$
How would you prove it? The proof certainly is not obvious as far as I can tell.
$endgroup$
– Eric Wofsey
Jan 9 at 3:20
$begingroup$
How would you prove it? The proof certainly is not obvious as far as I can tell.
$endgroup$
– Eric Wofsey
Jan 9 at 3:20
$begingroup$
if alternate interior angles are congruent then alternate exterior angles are congruent (as they are complementary to the interior). Then using straight angles and transitivity we can show that the same side interior/exterior angles are congruent.
$endgroup$
– Vasya
Jan 9 at 3:41
$begingroup$
if alternate interior angles are congruent then alternate exterior angles are congruent (as they are complementary to the interior). Then using straight angles and transitivity we can show that the same side interior/exterior angles are congruent.
$endgroup$
– Vasya
Jan 9 at 3:41
$begingroup$
And how do you conclude that the lines are parallel?
$endgroup$
– Eric Wofsey
Jan 9 at 3:51
$begingroup$
And how do you conclude that the lines are parallel?
$endgroup$
– Eric Wofsey
Jan 9 at 3:51
|
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