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On triangle $ABC$, with angles $\alpha$ over $A$, $\beta$ over $B$, and $\gamma$ over $C$. Where $\gamma$ is $140^\circ$.

On $\overline{AB}$ lies point $D$ (different from $A$ and $B$).

On $\overline{AC}$ lies point $E$ (different from $A$ and $C$).

$\overline{AE}$, $\overline{ED}$, $\overline{DC}$, $\overline{CB}$ 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 $180^\circ$.

enter image description here

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    $\begingroup$ What have you done? Have you, for instance, drawn this thing? $\endgroup$ – Arthur Mar 30 at 17:16
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    $\begingroup$ 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
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    $\begingroup$ 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
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I drew a simple diagram from which one can deduce that $$\alpha=10^\circ$$ $$\beta=40^\circ-\alpha=30^\circ$$ by noticing that angle $DEC=2\alpha$ and hence angle $DCB=140^\circ-2\alpha$ giving the value of the angle $DBC=20^\circ+\alpha$. But as angle $DBC$ is equal to angle $ABC$ we get that $$40^\circ-\alpha=20^\circ+\alpha\implies \alpha=10^\circ$$

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  • $\begingroup$ 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
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    $\begingroup$ $ADE=\alpha$ due to the isosceles triangle formed. $AED=180^\circ-2\alpha$ as angles in a triangle add to $180^\circ$. $DEC=180^\circ-AED=2\alpha$ 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
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The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...

And $$\angle ACB\not= \angle DCA$$

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Ok. Got it. Thanks for the picture. $\endgroup$ – Peter Parada Mar 30 at 17:40

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