What is the value of $alpha$ and $beta$ in a triangle? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 11:00UTC (8:00pm US/Eastern)Proof that the angle sum of a triangle is always greater than 180 degrees in elliptic geometryWhat characteristic of the triangle leads the the existence of the orthocenter$sinalpha+sinbeta+singamma>2$ Where $alpha$, $beta$ and $gamma$ are angles from an acute-angled triangle.Show that for triangle ABC, with complex numbers for the coordinates, that we have the following equationProving $cscfracalpha2+cscfracbeta2+cscfracgamma2 ge 6$, where $alpha$, $beta$, $gamma$ are the angles of a trianglePoint in a triangle plane determining any anglesGiven sides and a bisection, find angles in a triangleHow to find triangle height if I know its area and anglesFind angle given two isosceles triangles.Solving for the value of $ angle CEB$ - $frac14$ $angle CBA$ where $E$ is an exterior point of $triangle ABC$
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What is the value of $alpha$ and $beta$ in a triangle?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 11:00UTC (8:00pm US/Eastern)Proof that the angle sum of a triangle is always greater than 180 degrees in elliptic geometryWhat characteristic of the triangle leads the the existence of the orthocenter$sinalpha+sinbeta+singamma>2$ Where $alpha$, $beta$ and $gamma$ are angles from an acute-angled triangle.Show that for triangle ABC, with complex numbers for the coordinates, that we have the following equationProving $cscfracalpha2+cscfracbeta2+cscfracgamma2 ge 6$, where $alpha$, $beta$, $gamma$ are the angles of a trianglePoint in a triangle plane determining any anglesGiven sides and a bisection, find angles in a triangleHow to find triangle height if I know its area and anglesFind angle given two isosceles triangles.Solving for the value of $ angle CEB$ - $frac14$ $angle CBA$ where $E$ is an exterior point of $triangle ABC$
$begingroup$
On triangle $ABC$, with angles $alpha$ over $A$, $beta$ over $B$, and $gamma$ over $C$. Where $gamma$ is $140^circ$.
On $overlineAB$ lies point $D$ (different from $A$ and $B$).
On $overlineAC$ lies point $E$ (different from $A$ and $C$).
$overlineAE$, $overlineED$, $overlineDC$, $overlineCB$ are of same length.
What is the value of α and β?
EDIT: Here's my "solution", however the sum of the angles is greater than $180^circ$.
geometry triangles
edited Mar 31 at 2:08
Rócherz
3,0263823
asked Mar 30 at 17:13
Peter ParadaPeter Parada
1348
$endgroup$
add a comment |
$begingroup$
On triangle $ABC$, with angles $alpha$ over $A$, $beta$ over $B$, and $gamma$ over $C$. Where $gamma$ is $140^circ$.
On $overlineAB$ lies point $D$ (different from $A$ and $B$).
On $overlineAC$ lies point $E$ (different from $A$ and $C$).
$overlineAE$, $overlineED$, $overlineDC$, $overlineCB$ are of same length.
What is the value of α and β?
EDIT: Here's my "solution", however the sum of the angles is greater than $180^circ$.
geometry triangles
edited Mar 31 at 2:08
Rócherz
3,0263823
asked Mar 30 at 17:13
Peter ParadaPeter Parada
1348
$endgroup$
1
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
Mar 30 at 17:16
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
Mar 30 at 17:18
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
Mar 30 at 17:38
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
Mar 30 at 21:59
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
Mar 30 at 22:08
add a comment |
$begingroup$
On triangle $ABC$, with angles $alpha$ over $A$, $beta$ over $B$, and $gamma$ over $C$. Where $gamma$ is $140^circ$.
On $overlineAB$ lies point $D$ (different from $A$ and $B$).
On $overlineAC$ lies point $E$ (different from $A$ and $C$).
$overlineAE$, $overlineED$, $overlineDC$, $overlineCB$ are of same length.
What is the value of α and β?
EDIT: Here's my "solution", however the sum of the angles is greater than $180^circ$.
geometry triangles
edited Mar 31 at 2:08
Rócherz
3,0263823
asked Mar 30 at 17:13
Peter ParadaPeter Parada
1348
$endgroup$
On triangle $ABC$, with angles $alpha$ over $A$, $beta$ over $B$, and $gamma$ over $C$. Where $gamma$ is $140^circ$.
On $overlineAB$ lies point $D$ (different from $A$ and $B$).
On $overlineAC$ lies point $E$ (different from $A$ and $C$).
$overlineAE$, $overlineED$, $overlineDC$, $overlineCB$ are of same length.
What is the value of α and β?
EDIT: Here's my "solution", however the sum of the angles is greater than $180^circ$.
geometry triangles
geometry triangles
edited Mar 31 at 2:08
Rócherz
3,0263823
asked Mar 30 at 17:13
Peter ParadaPeter Parada
1348
edited Mar 31 at 2:08
Rócherz
3,0263823
asked Mar 30 at 17:13
Peter ParadaPeter Parada
1348
edited Mar 31 at 2:08
Rócherz
3,0263823
edited Mar 31 at 2:08
Rócherz
3,0263823
edited Mar 31 at 2:08
Rócherz
3,0263823
3,0263823
asked Mar 30 at 17:13
Peter ParadaPeter Parada
1348
asked Mar 30 at 17:13
Peter ParadaPeter Parada
1348
asked Mar 30 at 17:13
Peter ParadaPeter Parada
1348
1348
1
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
Mar 30 at 17:16
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
Mar 30 at 17:18
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
Mar 30 at 17:38
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
Mar 30 at 21:59
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
Mar 30 at 22:08
add a comment |
1
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
Mar 30 at 17:16
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
Mar 30 at 17:18
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
Mar 30 at 17:38
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
Mar 30 at 21:59
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
Mar 30 at 22:08
1
1
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
Mar 30 at 17:16
$begingroup$
What have you done? Have you, for instance, drawn this thing?
$endgroup$
– Arthur
Mar 30 at 17:16
1
1
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
Mar 30 at 17:18
$begingroup$
I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
$endgroup$
– Peter Parada
Mar 30 at 17:18
3
3
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
Mar 30 at 17:38
$begingroup$
Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
$endgroup$
– TonyK
Mar 30 at 17:38
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
Mar 30 at 21:59
$begingroup$
If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
$endgroup$
– MackTuesday
Mar 30 at 21:59
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
Mar 30 at 22:08
$begingroup$
It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
$endgroup$
– MackTuesday
Mar 30 at 22:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered Mar 30 at 17:32
Peter ForemanPeter Foreman
7,1741318
$endgroup$
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
Mar 30 at 17:45
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
Mar 30 at 17:53
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
Mar 30 at 17:58
add a comment |
$begingroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered Mar 30 at 17:32
Dr. MathvaDr. Mathva
3,493630
$endgroup$
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
Mar 30 at 17:35
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
Mar 30 at 17:38
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
Mar 30 at 17:40
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered Mar 30 at 17:32
Peter ForemanPeter Foreman
7,1741318
$endgroup$
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
Mar 30 at 17:45
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
Mar 30 at 17:53
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
Mar 30 at 17:58
add a comment |
$begingroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered Mar 30 at 17:32
Peter ForemanPeter Foreman
7,1741318
$endgroup$
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
Mar 30 at 17:45
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
Mar 30 at 17:53
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
Mar 30 at 17:58
add a comment |
$begingroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered Mar 30 at 17:32
Peter ForemanPeter Foreman
7,1741318
$endgroup$
I drew a simple diagram from which one can deduce that
$$alpha=10^circ$$
$$beta=40^circ-alpha=30^circ$$
by noticing that angle $DEC=2alpha$ and hence angle $DCB=140^circ-2alpha$ giving the value of the angle $DBC=20^circ+alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that
$$40^circ-alpha=20^circ+alphaimplies alpha=10^circ$$
answered Mar 30 at 17:32
Peter ForemanPeter Foreman
7,1741318
answered Mar 30 at 17:32
Peter ForemanPeter Foreman
7,1741318
answered Mar 30 at 17:32
Peter ForemanPeter Foreman
7,1741318
answered Mar 30 at 17:32
Peter ForemanPeter Foreman
7,1741318
7,1741318
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
Mar 30 at 17:45
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
Mar 30 at 17:53
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
Mar 30 at 17:58
add a comment |
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
Mar 30 at 17:45
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
Mar 30 at 17:53
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
Mar 30 at 17:58
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
$begingroup$
Nice.......[+1]
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
Mar 30 at 17:45
$begingroup$
@peter-foreman How have you deduced that DEC is 2𝛼?
$endgroup$
– Peter Parada
Mar 30 at 17:45
1
1
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
Mar 30 at 17:53
$begingroup$
$ADE=alpha$ due to the isosceles triangle formed. $AED=180^circ-2alpha$ as angles in a triangle add to $180^circ$. $DEC=180^circ-AED=2alpha$ as a straight line contains $180^circ$.
$endgroup$
– Peter Foreman
Mar 30 at 17:53
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
Mar 30 at 17:58
$begingroup$
Thank you. Very nice!
$endgroup$
– Peter Parada
Mar 30 at 17:58
add a comment |
$begingroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered Mar 30 at 17:32
Dr. MathvaDr. Mathva
3,493630
$endgroup$
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
Mar 30 at 17:35
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
Mar 30 at 17:38
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
Mar 30 at 17:40
add a comment |
$begingroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered Mar 30 at 17:32
Dr. MathvaDr. Mathva
3,493630
$endgroup$
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
Mar 30 at 17:35
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
Mar 30 at 17:38
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
Mar 30 at 17:40
add a comment |
$begingroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered Mar 30 at 17:32
Dr. MathvaDr. Mathva
3,493630
$endgroup$
The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...
And $$angle ACBnot= angle DCA$$
answered Mar 30 at 17:32
Dr. MathvaDr. Mathva
3,493630
edited Mar 30 at 17:39
answered Mar 30 at 17:32
Dr. MathvaDr. Mathva
3,493630
answered Mar 30 at 17:32
Dr. MathvaDr. Mathva
3,493630
answered Mar 30 at 17:32
Dr. MathvaDr. Mathva
3,493630
3,493630
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
Mar 30 at 17:35
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
Mar 30 at 17:38
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
Mar 30 at 17:40
add a comment |
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
$endgroup$
– Peter Parada
Mar 30 at 17:35
1
$begingroup$
No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
$endgroup$
– Peter Foreman
Mar 30 at 17:38
$begingroup$
Exactly @PeterForeman!
$endgroup$
– Dr. Mathva
Mar 30 at 17:39
1
$begingroup$
Ok. Got it. Thanks for the picture.
$endgroup$
– Peter Parada
Mar 30 at 17:40
$begingroup$
Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
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– Peter Parada
Mar 30 at 17:35
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Not sure if we are on the same page, but I am assuming exacly that. If 2 sides are equal, then 2 angles in triangle are also equal. - "The two angles opposite the legs are equal and are always acute" - en.wikipedia.org/wiki/Isosceles_triangle
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– Peter Parada
Mar 30 at 17:35
1
1
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No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
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– Peter Foreman
Mar 30 at 17:38
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No what he means is that all of the triangles in your diagram have different valued base angles - they are not all simply $x$. Just because two side lengths are the same does not imply all angles of a triangle are the same.
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– Peter Foreman
Mar 30 at 17:38
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Exactly @PeterForeman!
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– Dr. Mathva
Mar 30 at 17:39
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Exactly @PeterForeman!
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– Dr. Mathva
Mar 30 at 17:39
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Ok. Got it. Thanks for the picture.
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– Peter Parada
Mar 30 at 17:40
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Ok. Got it. Thanks for the picture.
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– Peter Parada
Mar 30 at 17:40
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What have you done? Have you, for instance, drawn this thing?
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– Arthur
Mar 30 at 17:16
1
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I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination.
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– Peter Parada
Mar 30 at 17:18
3
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Your drawing is hopelessly inaccurate. For example, $x$ (which should be $alpha$, by the way) seems to range from $40^circ$ to $110^circ$. Surely you can do better than that?
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– TonyK
Mar 30 at 17:38
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If $x+y neq 180$, points $A$, $E$, and $C$ are not collinear.
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– MackTuesday
Mar 30 at 21:59
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It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$.
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– MackTuesday
Mar 30 at 22:08