Geometrical problem in Newton's “Principia”.
Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, RP2 (that is, QR x RL) will be to QT2 as AV2 to PV2.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
geometry euclidean-geometry
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
335
add a comment |
Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, RP2 (that is, QR x RL) will be to QT2 as AV2 to PV2.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
geometry euclidean-geometry
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
335
1
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
– Blue
Jan 6 at 18:44
add a comment |
Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, RP2 (that is, QR x RL) will be to QT2 as AV2 to PV2.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
geometry euclidean-geometry
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
335
Let VQPA be the circumference
of the circle, S the given
point toward which the force
tends as to its center, P the body
revolving in the circumference, Q
the place to which it will move
next, and PRZ the tangent of
the circle at the previous place.
Through point S draw chord PV;
and when the diameter VA of
the circle has been drawn, join
AP; and to SP drop perpendicular
QT, which when produced meets the tangent PR at Z; and finally through
point Q draw LR parallel to SP and meeting both the circle at L and the
tangent PZ at R. Then because the triangles ZQR, ZTP, and VPA are
similar, RP2 (that is, QR x RL) will be to QT2 as AV2 to PV2.
My question is how can we prove the similarity of triangles ZQR, ZTP, and VPA?
geometry euclidean-geometry
geometry euclidean-geometry
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
335
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
335
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
335
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
335
asked Jan 6 at 17:43
Vaggelis KyrilasVaggelis Kyrilas
335
335
1
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
– Blue
Jan 6 at 18:44
add a comment |
1
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
– Blue
Jan 6 at 18:44
1
1
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
– Blue
Jan 6 at 18:44
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
– Blue
Jan 6 at 18:44
add a comment |
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1
$triangle ZQRsim triangle ZTP$ because $overline{QR}paralleloverline{TP}$. For the other similarity, first note that $angle VPA$ is a right angle (via Thales' Theorem). Also, if we introduce $O$ as the center of the circle (aka, the midpoint of $overline{VA}$), we can do a little angle-chasing to show $$angle ZPT=90^circ−angle OPT=angle OPA = angle A$$ Thus, $triangle ZTPsimtriangle VPA$ by Angle-Angle Similarity.
– Blue
Jan 6 at 18:44