MHB Geometry Challenge: Prove $\angle ADE=\angle BDC$ in Convex Quadrilateral $ADBE$

[anemone]

In convex quadrilateral $ADBE$, there is a point $C$ within $\triangle ABE$ such that $\angle EAD+\angle CAB=\angle EBD+\angle CBA=180^{\circ}$.

Prove that $\angle ADE=\angle BDC$.

 

[anemone]

Spoiler: Solution of other

[TIKZ]
\begin{scope}
\draw (0,0) circle(3);
\end{scope}
\coordinate[label=left: E] (E) at (-3,0);
\coordinate[label=below: D] (D) at (1.2,-2.75);
\coordinate[label=below: A] (A) at (-1,.-2.828);
\coordinate[label=right: F] (F) at (2.9,-0.768);
\coordinate[label=above: B] (B) at (-1,-0.26);
\coordinate[label=above: C] (C) at (-2,-1.1);
\draw (A) -- (E);
\draw (A) -- (D);
\draw (F) -- (D);
\draw (E) -- (F);
\draw (A) -- (B);
\draw (A) -- (F);
\draw (E) -- (D);
\draw (B) -- (D);
\draw [dashed] (C) -- (D);
\draw [dashed] (C) -- (B);
\draw [dashed] (C) -- (A);
[/TIKZ]

Let F be the second intersection of the circumcircle of $\triangle EAD$ and line $EB$. Then $\angle DBF=180^{\circ}-\angle EBD=\angle CBA$. Moreover,

$\begin{align*}\angle BDF&=180^{\circ}-\angle AEB-\angle ADB\\&=180^{\circ}-(360^{\circ}-\angle EAD-\angle EBD)\\&= 180^{\circ}-(\angle CAB+\angle CBA)\\&=\angle BCA\end{align*}$

These two relations give $\angle BDF \simeq \triangle BCA$.

So $\dfrac{BD}{BF}=\dfrac{BC}{BA}$ Together with $\angle DBF=\angle CBA$, we have $\triangle BDC \simeq \triangle BFA$.

This results in $\angle ADE=\angle AFE=\angle BFA=\angle BDC$. (Q.E.D.)

 

[MarkFL]

You always do such a nice job with your presentations...the TiKZ drawings are really nice (and add such quality), and I know they take some effort too. :)

 

[anemone]

Mark, to be completely honest, I have to say once you get to know some simple commands like how to draw a circle, joining lines, labeling angles, coloring some region, etc, then basically you can draw anything out of these simple commands. Of course, my other trick is always look for Klaas for help when I got stuck in some effect I want to produce to my diagram, hehehe... (Happy)