Can You Find the Angle in This Isosceles Triangle Problem?

[anemone]

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 $EDC$.

 

[Opalg]

Spoiler: My solution

[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 $70^\circ$.