# [SOLVED]Geometry challenge

#### anemone

##### MHB POTW Director
Staff member
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

##### MHB POTW Director
Staff member
\begin{tikzpicture}

\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);

\end{tikzpicture}

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.)