Proof with d-quadrilateral
$begingroup$
This is the question that I am working on currently I am just stumped on how to go about solving it. I don't really know how I can use the Euclidean triangle to solve it either. Any advice or tips would be much appreciated. Thank you very much.
geometry noneuclidean-geometry
asked Nov 30 '18 at 22:06
MosephMoseph
325
$endgroup$
add a comment |
$begingroup$
This is the question that I am working on currently I am just stumped on how to go about solving it. I don't really know how I can use the Euclidean triangle to solve it either. Any advice or tips would be much appreciated. Thank you very much.
geometry noneuclidean-geometry
asked Nov 30 '18 at 22:06
MosephMoseph
325
$endgroup$
add a comment |
$begingroup$
This is the question that I am working on currently I am just stumped on how to go about solving it. I don't really know how I can use the Euclidean triangle to solve it either. Any advice or tips would be much appreciated. Thank you very much.
geometry noneuclidean-geometry
asked Nov 30 '18 at 22:06
MosephMoseph
325
$endgroup$
This is the question that I am working on currently I am just stumped on how to go about solving it. I don't really know how I can use the Euclidean triangle to solve it either. Any advice or tips would be much appreciated. Thank you very much.
geometry noneuclidean-geometry
geometry noneuclidean-geometry
asked Nov 30 '18 at 22:06
MosephMoseph
325
asked Nov 30 '18 at 22:06
MosephMoseph
325
asked Nov 30 '18 at 22:06
MosephMoseph
325
asked Nov 30 '18 at 22:06
MosephMoseph
325
asked Nov 30 '18 at 22:06
MosephMoseph
325
325
add a comment |
add a comment |
1 Answer
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That is Poincare's hyperbolic disk, correct? I'll simplify notation a little bit, and call $A = angle EAB$, $D = angle EDC$, $P = angle DPA$, all angles in the hyperbolic plane - which by conformity equal the angles formed by the intersecting $d$-lines seen as curves in the Euclidean plane where the hyperbolic plane is modeled by Poincare's disk. The angle $angle EDP$ is $pi - D$, so you are asked to prove $A ne pi - D$.
If you join $D$ and $P$ with a straight line then you get the Euclidean triangle $triangle APD$. Call $D' = angle ADP$, and $P' = angle APD$ in the Euclidean triangle. Since $P$ is right at the boundary of the Euclidean disk then $P = 0$ (all the d-lines enter the boundary of the disk perpendicularly, so they form a null angle). We also have $D < D'$ and $0 < P'$, hence $A + D + P = A + D < A + D' + P' = pi$, and from here $A < pi - D$, which implies the statement being proved (and at the same time shows how Euclid's 5th postulate fails in hyperbolic geometry, since $A+D < pi$ and the d-lines $AP$ and $DP$ do no meet).
answered Dec 1 '18 at 1:19
mlerma54mlerma54
1,162148
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1 Answer
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$begingroup$
That is Poincare's hyperbolic disk, correct? I'll simplify notation a little bit, and call $A = angle EAB$, $D = angle EDC$, $P = angle DPA$, all angles in the hyperbolic plane - which by conformity equal the angles formed by the intersecting $d$-lines seen as curves in the Euclidean plane where the hyperbolic plane is modeled by Poincare's disk. The angle $angle EDP$ is $pi - D$, so you are asked to prove $A ne pi - D$.
If you join $D$ and $P$ with a straight line then you get the Euclidean triangle $triangle APD$. Call $D' = angle ADP$, and $P' = angle APD$ in the Euclidean triangle. Since $P$ is right at the boundary of the Euclidean disk then $P = 0$ (all the d-lines enter the boundary of the disk perpendicularly, so they form a null angle). We also have $D < D'$ and $0 < P'$, hence $A + D + P = A + D < A + D' + P' = pi$, and from here $A < pi - D$, which implies the statement being proved (and at the same time shows how Euclid's 5th postulate fails in hyperbolic geometry, since $A+D < pi$ and the d-lines $AP$ and $DP$ do no meet).
answered Dec 1 '18 at 1:19
mlerma54mlerma54
1,162148
$endgroup$
add a comment |
$begingroup$
That is Poincare's hyperbolic disk, correct? I'll simplify notation a little bit, and call $A = angle EAB$, $D = angle EDC$, $P = angle DPA$, all angles in the hyperbolic plane - which by conformity equal the angles formed by the intersecting $d$-lines seen as curves in the Euclidean plane where the hyperbolic plane is modeled by Poincare's disk. The angle $angle EDP$ is $pi - D$, so you are asked to prove $A ne pi - D$.
If you join $D$ and $P$ with a straight line then you get the Euclidean triangle $triangle APD$. Call $D' = angle ADP$, and $P' = angle APD$ in the Euclidean triangle. Since $P$ is right at the boundary of the Euclidean disk then $P = 0$ (all the d-lines enter the boundary of the disk perpendicularly, so they form a null angle). We also have $D < D'$ and $0 < P'$, hence $A + D + P = A + D < A + D' + P' = pi$, and from here $A < pi - D$, which implies the statement being proved (and at the same time shows how Euclid's 5th postulate fails in hyperbolic geometry, since $A+D < pi$ and the d-lines $AP$ and $DP$ do no meet).
answered Dec 1 '18 at 1:19
mlerma54mlerma54
1,162148
$endgroup$
add a comment |
$begingroup$
That is Poincare's hyperbolic disk, correct? I'll simplify notation a little bit, and call $A = angle EAB$, $D = angle EDC$, $P = angle DPA$, all angles in the hyperbolic plane - which by conformity equal the angles formed by the intersecting $d$-lines seen as curves in the Euclidean plane where the hyperbolic plane is modeled by Poincare's disk. The angle $angle EDP$ is $pi - D$, so you are asked to prove $A ne pi - D$.
If you join $D$ and $P$ with a straight line then you get the Euclidean triangle $triangle APD$. Call $D' = angle ADP$, and $P' = angle APD$ in the Euclidean triangle. Since $P$ is right at the boundary of the Euclidean disk then $P = 0$ (all the d-lines enter the boundary of the disk perpendicularly, so they form a null angle). We also have $D < D'$ and $0 < P'$, hence $A + D + P = A + D < A + D' + P' = pi$, and from here $A < pi - D$, which implies the statement being proved (and at the same time shows how Euclid's 5th postulate fails in hyperbolic geometry, since $A+D < pi$ and the d-lines $AP$ and $DP$ do no meet).
answered Dec 1 '18 at 1:19
mlerma54mlerma54
1,162148
$endgroup$
That is Poincare's hyperbolic disk, correct? I'll simplify notation a little bit, and call $A = angle EAB$, $D = angle EDC$, $P = angle DPA$, all angles in the hyperbolic plane - which by conformity equal the angles formed by the intersecting $d$-lines seen as curves in the Euclidean plane where the hyperbolic plane is modeled by Poincare's disk. The angle $angle EDP$ is $pi - D$, so you are asked to prove $A ne pi - D$.
If you join $D$ and $P$ with a straight line then you get the Euclidean triangle $triangle APD$. Call $D' = angle ADP$, and $P' = angle APD$ in the Euclidean triangle. Since $P$ is right at the boundary of the Euclidean disk then $P = 0$ (all the d-lines enter the boundary of the disk perpendicularly, so they form a null angle). We also have $D < D'$ and $0 < P'$, hence $A + D + P = A + D < A + D' + P' = pi$, and from here $A < pi - D$, which implies the statement being proved (and at the same time shows how Euclid's 5th postulate fails in hyperbolic geometry, since $A+D < pi$ and the d-lines $AP$ and $DP$ do no meet).
answered Dec 1 '18 at 1:19
mlerma54mlerma54
1,162148
answered Dec 1 '18 at 1:19
mlerma54mlerma54
1,162148
answered Dec 1 '18 at 1:19
mlerma54mlerma54
1,162148
answered Dec 1 '18 at 1:19
mlerma54mlerma54
1,162148
1,162148
add a comment |
add a comment |
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