MHB How Long is Side CD in a Special Trapezium?

[anemone]

Here is this week's POTW:

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$ABCD$ is a trapezium with $AD$ is parallel to $BC$. Given that $BC=BD=1,\, AB=AC$, $CD<1$ and $\angle BAC+\angle BDC=180^\circ$, find $CD$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. Opalg

Solution from Opalg:

Spoiler

[TIKZ]\coordinate [label=below left: $A$] (A) at (5,-4.05) ;
\coordinate [label=above: $B$] (B) at (0,0) ;
\coordinate [label=above: $C$] (C) at (10,0) ;
\coordinate [label=below: $D$] (D) at (9.14,-4.05) ;
\draw (A) -- (B) -- node[above] {$1$} (C) -- (D) -- (A) -- (C) ;
\draw (B) -- node[above] {$1$} (D) ;
\node at (5,-3.7) {$\alpha$};
\node at (9,-3.7) {$\beta$};
\node at (5.5,-3.9) {$\gamma$};
\node at (9.5,-0.2) {$\gamma$};
\node at (9.7,-0.5) {$\gamma$};
\node at (0.7,-0.3) {$\gamma$};[/TIKZ]

Let $\alpha = \angle BAC$ and $\beta = \angle BDC$, so that $\alpha + \beta = 180^\circ$. Also, let $\gamma = \frac12\beta$.

In the isosceles triangle $ABC$, the angle at $A$ is $\alpha$, so the other two angles must be $\frac12(180^\circ - \alpha) = \gamma$. Since $\angle BCD = \beta$, it follows that $\angle ACD = \gamma$.

Since $BC$ and $AD$ are parallel, $\angle CAD = \angle BCA = \gamma$. Thus the triangle $ACD$ is isosceles and therefore $AD = CD$.

From the isosceles triangle $BCD$ it follows that $AD = CD = 2\cos\beta$.

Now take coordinates with $B$ at the origin and $C = (1,0)$. The $x$-coordinate of $A$ is $\frac12$, and the $x$-coordinate of $D$ is $1 - CD\cos\beta = 1 - 2\cos^2\beta$. But $A$ and $D$ have the same $y$-coordinate and so $AD = \frac12 - 2\cos^2\beta$.

Therefore $ \frac12 - 2\cos^2\beta = 2\cos\beta$, from which $4\cos^2\beta + 4\cos\beta - 1 = 0$, a quadratic with solution $\cos\beta = \frac12(\sqrt2 - 1)$ (the other solution is less than $-1$, so cannot be a cosine).

Hence $CD = 2\cos\beta = \sqrt2 - 1$.