MHB Can You Find the Angle in This Isosceles Triangle Problem?

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The discussion revolves around a geometric problem involving an isosceles triangle ABC with a base angle of 80 degrees. A point D is located on side AB such that AD equals BC, and point E is on ray CB such that AC equals EC. Participants are encouraged to solve the problem and share their solutions. The original poster expresses interest in the challenge but has not yet found a solution. The problem invites collaborative problem-solving within the forum.
anemone
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Hello MHB, I saw one question that really tickles my intellectual fancy and because of the limited spare time that I have, I could not say I have solved it already! But, I will most definitely give the question more thought and will post back if I find a good solution to it.

Here goes the question...and if this question intrigues you, please feel free to try it and in case you have solved it, please share it with us!

In the isosceles triangle $ABC$, the angle at the base $BC$ is equal to $80^{\circ}$. On the side $AB$ the point $D$ is chosen such that $AD=BC$ and on the ray $CB$ the point $E$ is chosen such that $AC=EC$. Find the angle $EDC$.
 
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[TIKZ]\coordinate [label=left:$A$] (A) at (80:9) ;
\coordinate [label=below:$B$] (B) at (0,0) ;
\coordinate [label=below:$C$] (C) at (3,0) ;
\coordinate [label=left:$D$] (D) at (80:6) ;
\coordinate [label=below:$E$] (E) at (-6,0) ;
\draw (B) -- node[ below right ]{$x$}(A) -- node[ below right ]{$x$}(C) -- node[ below ]{$x$}(E) -- node[ above ]{$y$}(D) -- (C) ;
\draw (-5.25,0.25) node{$40^\circ$} ;
\draw (0.5,0.25) node{$80^\circ$} ;
\draw (0.5,5) node{$40^\circ$} ;[/TIKZ]
Let $x = AB = AC = EC$ and let $y = ED$.

In the isosceles triangle $ABC$, $BC = 2x\cos80^\circ$. Since $AD = BC$, it follows that $EB = DB = x(1 - 2\cos80^\circ)$. Thus the triangle $DBE$ is isosceles, and since the apex angle at $B$ is $100^\circ$ it follows that the base angles $DEB$ and $EDB$ are both $40^\circ$.

By the sine rule in the triangle $DBE$, $\dfrac{DE}{\sin100^\circ} = \dfrac{EB}{\sin40^\circ}$, or $\dfrac{y}{\sin80^\circ} = \dfrac{x(1 - 2\cos80^\circ)}{\sin40^\circ}$. Therefore $$y\sin40^\circ = x\sin80^\circ(1 - 2\cos80^\circ) = 2x\sin40^\circ\cos40^\circ\bigl(1 - 2(2\cos^240^\circ - 1)\bigr),$$ $$y = 2x\cos40^\circ(3 - 4\cos^240^\circ) = -2x(4\cos^340^\circ - 3\cos40^\circ).$$ The formula $\cos3\theta = 4\cos^3\theta - 3\cos\theta$ shows that $4\cos^340^\circ - 3\cos40^\circ = \cos120^\circ = -\cos60^\circ = -\frac12$. Therefore $y = -2x\bigl(-\frac12\bigr) = x$. So $ED = EC$ and the triangle $EDC$ is isosceles. The angle at the apex $E$ is $40^\circ$ and so the base angles (one of which is $EDC$) are $70^\circ$.
 
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