Proving the diagonals of a quadrilateral are equal This is an easy question but it is troubling me a lot:
$ABDC$ is a convex quadrilateral, with $AB=BC=AC$ and $\angle BDC=150^{\circ}$. Show that its diagonals are equal. I have tried fiddling with the cosine rule, but it hasn't worked. Please help!

EDIT: I don't want to use vectors, only standard geometry and trigonometry.
 A: Converting comment to answer with a few more details, as requested.

Draw the circumcircle of $\triangle BCD$; let its center be $K$, and let $D^\prime$ be a point on the major arc $\stackrel{\frown}{BC}$. Note that $\angle BDC$ and $\angle BD^\prime C$ are supplementary. By the Inscribed Angle Theorem, point $K$ is such that 
$$\angle BKC = 2\;\angle BD^\prime C = 2\;( 180^\circ - \angle BDC ) = 60^\circ$$
Point $K$ is also necessarily on the perpendicular bisector of $\overline{BC}$. These two facts are enough to determine $K$ uniquely (why?), and since those facts are true of point $A$, we have that $A=K$. Thus, $\overline{AB}$, $\overline{AC}$, and $\overline{AD}$ are all congruent, by virtue of $A$ being the center of the circumcircle of $\triangle BCD$; and then the diagonals $\overline{BC}$ and $\overline{AD}$ are congruent, by virtue of $\triangle ABC$ being equilateral. 
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Let's $\ds{\verts{\vec{\rm AB}} = \verts{\vec{\rm BC}} = \verts{\vec{\rm AC}}
           \equiv \ell}$

$$
\ell\
\overbrace{\sin\pars{60^{o}}}^{\ds{\root{3}/2}} 
+ {\ell/2 \over \underbrace{\tan\pars{75^{o}}}_{2 + \root{3}}}
=
\pars{{\root{3} \over 2} + \half\,{\root{3} - 2 \over 3 - 4}}\,\ell
={\root{3} - \pars{\root{3} - 2} \over 2}\,\ell = \color{#00f}{\Large\ell}
$$

