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

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- Feb 14, 2012

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

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- Feb 14, 2012

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\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.)

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