Geometric Proof: Two parallel lines in circle, prove congruent arcs.
$begingroup$
In this geometric proof sheet, there is a circle with two parallel lines (one a radius and the other a chord, with an endpoint on the diameter of the circle.
You can see a diagram here (question 6 page 102: https://www.engageny.org/resource/geometry-module-5-topic-b-lesson-8/file/127991)
The solution does not make sense to me, specifically the part where it claims that angle AED equals angle EAD through substitution. Help please!
geometry
edited Feb 26 '16 at 23:50
Brian Tung
26.2k32656
asked Feb 26 '16 at 13:05
HaimHaim
380210
$endgroup$
add a comment |
$begingroup$
In this geometric proof sheet, there is a circle with two parallel lines (one a radius and the other a chord, with an endpoint on the diameter of the circle.
You can see a diagram here (question 6 page 102: https://www.engageny.org/resource/geometry-module-5-topic-b-lesson-8/file/127991)
The solution does not make sense to me, specifically the part where it claims that angle AED equals angle EAD through substitution. Help please!
geometry
edited Feb 26 '16 at 23:50
Brian Tung
26.2k32656
asked Feb 26 '16 at 13:05
HaimHaim
380210
$endgroup$
add a comment |
$begingroup$
In this geometric proof sheet, there is a circle with two parallel lines (one a radius and the other a chord, with an endpoint on the diameter of the circle.
You can see a diagram here (question 6 page 102: https://www.engageny.org/resource/geometry-module-5-topic-b-lesson-8/file/127991)
The solution does not make sense to me, specifically the part where it claims that angle AED equals angle EAD through substitution. Help please!
geometry
edited Feb 26 '16 at 23:50
Brian Tung
26.2k32656
asked Feb 26 '16 at 13:05
HaimHaim
380210
$endgroup$
In this geometric proof sheet, there is a circle with two parallel lines (one a radius and the other a chord, with an endpoint on the diameter of the circle.
You can see a diagram here (question 6 page 102: https://www.engageny.org/resource/geometry-module-5-topic-b-lesson-8/file/127991)
The solution does not make sense to me, specifically the part where it claims that angle AED equals angle EAD through substitution. Help please!
geometry
geometry
edited Feb 26 '16 at 23:50
Brian Tung
26.2k32656
asked Feb 26 '16 at 13:05
HaimHaim
380210
edited Feb 26 '16 at 23:50
Brian Tung
26.2k32656
asked Feb 26 '16 at 13:05
HaimHaim
380210
edited Feb 26 '16 at 23:50
Brian Tung
26.2k32656
edited Feb 26 '16 at 23:50
Brian Tung
26.2k32656
edited Feb 26 '16 at 23:50
Brian Tung
26.2k32656
26.2k32656
asked Feb 26 '16 at 13:05
HaimHaim
380210
asked Feb 26 '16 at 13:05
HaimHaim
380210
asked Feb 26 '16 at 13:05
HaimHaim
380210
380210
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The "substitution" refers to the fact that any angle in the sum may be replaced by a congruent angle. That is just algebra.
But the proof is incorrect. Nowhere do we are angle DEA congruent to any other angle before the algebraic substitution is made. So we cannot validly substitute for it.
A proper proof can be constructed using AE as an auxiliary segment. Triangle ACE is isosceles with base CE (the other two sides are radii of the circle), so angles ACE and AEC are congruent. AE is a transverse between parallel lines so alternating interior angles AEC and DAE are congruent. AC is a transversal between parallel lines so corresponding angles ACE and DAB are congruent. By the transitive property central angles DAE and DAB are congruent and so are their intercepted arcs.
answered Mar 3 '16 at 11:18
Oscar LanziOscar Lanzi
13.4k12136
$endgroup$
add a comment |
$begingroup$
The claim is correct, but the alleged proof is dead wrong. It so happens that in the figure all angles are about $60^circ$, which is misleading.
A simple proof is as follows: The central angle $alpha:=angle(BAD)$ is equal to the peripheral angle $angle(BCE)$, by assumption. It follows that the central angle $angle(BAE)$ corresponding to $angle(BCE)$ is $=2alpha$.
answered Aug 16 '17 at 12:10
Christian BlatterChristian Blatter
176k8115327
$endgroup$
add a comment |
$begingroup$
Refer to the figure:
$hspace{5cm}$
Line 1: draw the red line segments.
Line 2: note the equal radii $R$.
Line 3: the base angles $x,y,z$ are equal in each of the three triangles.
Line 5: the alternate angles are equal $angle AEC=angle EAD=x$.
Line 6: the sum of interior angles of $Delta AED$: $x+2y=180^circ$.
Line 7: (assuming $x=y$, which is not shown), $3y=180^circ$.
Alternative:
Line 7: the corresponding angles are equal: $angle ACE=angle BAD=x$.
Line 8: The triangles are congruent by SAS: $Delta ABDcong Delta ADE$.
Line 9: $BD=DE$.
Line 10: $widehat{BD}=widehat{DE}$.
answered Jun 15 '18 at 10:24
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The "substitution" refers to the fact that any angle in the sum may be replaced by a congruent angle. That is just algebra.
But the proof is incorrect. Nowhere do we are angle DEA congruent to any other angle before the algebraic substitution is made. So we cannot validly substitute for it.
A proper proof can be constructed using AE as an auxiliary segment. Triangle ACE is isosceles with base CE (the other two sides are radii of the circle), so angles ACE and AEC are congruent. AE is a transverse between parallel lines so alternating interior angles AEC and DAE are congruent. AC is a transversal between parallel lines so corresponding angles ACE and DAB are congruent. By the transitive property central angles DAE and DAB are congruent and so are their intercepted arcs.
answered Mar 3 '16 at 11:18
Oscar LanziOscar Lanzi
13.4k12136
$endgroup$
add a comment |
$begingroup$
The "substitution" refers to the fact that any angle in the sum may be replaced by a congruent angle. That is just algebra.
But the proof is incorrect. Nowhere do we are angle DEA congruent to any other angle before the algebraic substitution is made. So we cannot validly substitute for it.
A proper proof can be constructed using AE as an auxiliary segment. Triangle ACE is isosceles with base CE (the other two sides are radii of the circle), so angles ACE and AEC are congruent. AE is a transverse between parallel lines so alternating interior angles AEC and DAE are congruent. AC is a transversal between parallel lines so corresponding angles ACE and DAB are congruent. By the transitive property central angles DAE and DAB are congruent and so are their intercepted arcs.
answered Mar 3 '16 at 11:18
Oscar LanziOscar Lanzi
13.4k12136
$endgroup$
add a comment |
$begingroup$
The "substitution" refers to the fact that any angle in the sum may be replaced by a congruent angle. That is just algebra.
But the proof is incorrect. Nowhere do we are angle DEA congruent to any other angle before the algebraic substitution is made. So we cannot validly substitute for it.
A proper proof can be constructed using AE as an auxiliary segment. Triangle ACE is isosceles with base CE (the other two sides are radii of the circle), so angles ACE and AEC are congruent. AE is a transverse between parallel lines so alternating interior angles AEC and DAE are congruent. AC is a transversal between parallel lines so corresponding angles ACE and DAB are congruent. By the transitive property central angles DAE and DAB are congruent and so are their intercepted arcs.
answered Mar 3 '16 at 11:18
Oscar LanziOscar Lanzi
13.4k12136
$endgroup$
The "substitution" refers to the fact that any angle in the sum may be replaced by a congruent angle. That is just algebra.
But the proof is incorrect. Nowhere do we are angle DEA congruent to any other angle before the algebraic substitution is made. So we cannot validly substitute for it.
A proper proof can be constructed using AE as an auxiliary segment. Triangle ACE is isosceles with base CE (the other two sides are radii of the circle), so angles ACE and AEC are congruent. AE is a transverse between parallel lines so alternating interior angles AEC and DAE are congruent. AC is a transversal between parallel lines so corresponding angles ACE and DAB are congruent. By the transitive property central angles DAE and DAB are congruent and so are their intercepted arcs.
answered Mar 3 '16 at 11:18
Oscar LanziOscar Lanzi
13.4k12136
edited Mar 3 '16 at 11:47
answered Mar 3 '16 at 11:18
Oscar LanziOscar Lanzi
13.4k12136
answered Mar 3 '16 at 11:18
Oscar LanziOscar Lanzi
13.4k12136
answered Mar 3 '16 at 11:18
Oscar LanziOscar Lanzi
13.4k12136
13.4k12136
add a comment |
add a comment |
$begingroup$
The claim is correct, but the alleged proof is dead wrong. It so happens that in the figure all angles are about $60^circ$, which is misleading.
A simple proof is as follows: The central angle $alpha:=angle(BAD)$ is equal to the peripheral angle $angle(BCE)$, by assumption. It follows that the central angle $angle(BAE)$ corresponding to $angle(BCE)$ is $=2alpha$.
answered Aug 16 '17 at 12:10
Christian BlatterChristian Blatter
176k8115327
$endgroup$
add a comment |
$begingroup$
The claim is correct, but the alleged proof is dead wrong. It so happens that in the figure all angles are about $60^circ$, which is misleading.
A simple proof is as follows: The central angle $alpha:=angle(BAD)$ is equal to the peripheral angle $angle(BCE)$, by assumption. It follows that the central angle $angle(BAE)$ corresponding to $angle(BCE)$ is $=2alpha$.
answered Aug 16 '17 at 12:10
Christian BlatterChristian Blatter
176k8115327
$endgroup$
add a comment |
$begingroup$
The claim is correct, but the alleged proof is dead wrong. It so happens that in the figure all angles are about $60^circ$, which is misleading.
A simple proof is as follows: The central angle $alpha:=angle(BAD)$ is equal to the peripheral angle $angle(BCE)$, by assumption. It follows that the central angle $angle(BAE)$ corresponding to $angle(BCE)$ is $=2alpha$.
answered Aug 16 '17 at 12:10
Christian BlatterChristian Blatter
176k8115327
$endgroup$
The claim is correct, but the alleged proof is dead wrong. It so happens that in the figure all angles are about $60^circ$, which is misleading.
A simple proof is as follows: The central angle $alpha:=angle(BAD)$ is equal to the peripheral angle $angle(BCE)$, by assumption. It follows that the central angle $angle(BAE)$ corresponding to $angle(BCE)$ is $=2alpha$.
answered Aug 16 '17 at 12:10
Christian BlatterChristian Blatter
176k8115327
answered Aug 16 '17 at 12:10
Christian BlatterChristian Blatter
176k8115327
answered Aug 16 '17 at 12:10
Christian BlatterChristian Blatter
176k8115327
answered Aug 16 '17 at 12:10
Christian BlatterChristian Blatter
176k8115327
176k8115327
add a comment |
add a comment |
$begingroup$
Refer to the figure:
$hspace{5cm}$
Line 1: draw the red line segments.
Line 2: note the equal radii $R$.
Line 3: the base angles $x,y,z$ are equal in each of the three triangles.
Line 5: the alternate angles are equal $angle AEC=angle EAD=x$.
Line 6: the sum of interior angles of $Delta AED$: $x+2y=180^circ$.
Line 7: (assuming $x=y$, which is not shown), $3y=180^circ$.
Alternative:
Line 7: the corresponding angles are equal: $angle ACE=angle BAD=x$.
Line 8: The triangles are congruent by SAS: $Delta ABDcong Delta ADE$.
Line 9: $BD=DE$.
Line 10: $widehat{BD}=widehat{DE}$.
answered Jun 15 '18 at 10:24
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
$begingroup$
Refer to the figure:
$hspace{5cm}$
Line 1: draw the red line segments.
Line 2: note the equal radii $R$.
Line 3: the base angles $x,y,z$ are equal in each of the three triangles.
Line 5: the alternate angles are equal $angle AEC=angle EAD=x$.
Line 6: the sum of interior angles of $Delta AED$: $x+2y=180^circ$.
Line 7: (assuming $x=y$, which is not shown), $3y=180^circ$.
Alternative:
Line 7: the corresponding angles are equal: $angle ACE=angle BAD=x$.
Line 8: The triangles are congruent by SAS: $Delta ABDcong Delta ADE$.
Line 9: $BD=DE$.
Line 10: $widehat{BD}=widehat{DE}$.
answered Jun 15 '18 at 10:24
farruhotafarruhota
21.7k2842
$endgroup$
add a comment |
$begingroup$
Refer to the figure:
$hspace{5cm}$
Line 1: draw the red line segments.
Line 2: note the equal radii $R$.
Line 3: the base angles $x,y,z$ are equal in each of the three triangles.
Line 5: the alternate angles are equal $angle AEC=angle EAD=x$.
Line 6: the sum of interior angles of $Delta AED$: $x+2y=180^circ$.
Line 7: (assuming $x=y$, which is not shown), $3y=180^circ$.
Alternative:
Line 7: the corresponding angles are equal: $angle ACE=angle BAD=x$.
Line 8: The triangles are congruent by SAS: $Delta ABDcong Delta ADE$.
Line 9: $BD=DE$.
Line 10: $widehat{BD}=widehat{DE}$.
answered Jun 15 '18 at 10:24
farruhotafarruhota
21.7k2842
$endgroup$
Refer to the figure:
$hspace{5cm}$
Line 1: draw the red line segments.
Line 2: note the equal radii $R$.
Line 3: the base angles $x,y,z$ are equal in each of the three triangles.
Line 5: the alternate angles are equal $angle AEC=angle EAD=x$.
Line 6: the sum of interior angles of $Delta AED$: $x+2y=180^circ$.
Line 7: (assuming $x=y$, which is not shown), $3y=180^circ$.
Alternative:
Line 7: the corresponding angles are equal: $angle ACE=angle BAD=x$.
Line 8: The triangles are congruent by SAS: $Delta ABDcong Delta ADE$.
Line 9: $BD=DE$.
Line 10: $widehat{BD}=widehat{DE}$.
answered Jun 15 '18 at 10:24
farruhotafarruhota
21.7k2842
answered Jun 15 '18 at 10:24
farruhotafarruhota
21.7k2842
answered Jun 15 '18 at 10:24
farruhotafarruhota
21.7k2842
answered Jun 15 '18 at 10:24
farruhotafarruhota
21.7k2842
21.7k2842
add a comment |
add a comment |
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