# What is the value of $\alpha$ and $\beta$ in a triangle?

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 $$180^\circ$$.

• What have you done? Have you, for instance, drawn this thing? – Arthur Mar 30 at 17:16
• I had some solution where α and β were both 40° which was nonsense for me. Will do some drawing for better imagination. – Peter Parada Mar 30 at 17:18
• 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? – TonyK Mar 30 at 17:38
• If $x+y \neq 180$, points $A$, $E$, and $C$ are not collinear. – MackTuesday Mar 30 at 21:59
• It then follows that $AECB$ is a quadrilateral, not a triangle. It turns out $y = 0$. It all stems from the fact that $ED = CD$. – MackTuesday Mar 30 at 22:08

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 $$40^\circ-\alpha=20^\circ+\alpha\implies \alpha=10^\circ$$

• Nice.......[+1] – Dr. Mathva Mar 30 at 17:39
• @peter-foreman How have you deduced that DEC is 2𝛼? – Peter Parada Mar 30 at 17:45
• $ADE=\alpha$ 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 $180^\circ$. – Peter Foreman Mar 30 at 17:53
• Thank you. Very nice! – Peter Parada Mar 30 at 17:58

The problem is that you've assumed that two isosceles triangles with the same legs will have the same basis-angles...

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

• 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 – Peter Parada Mar 30 at 17:35
• 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. – Peter Foreman Mar 30 at 17:38
• Exactly @PeterForeman! – Dr. Mathva Mar 30 at 17:39
• Ok. Got it. Thanks for the picture. – Peter Parada Mar 30 at 17:40