# 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.

• The assumptions of the question appear to be inconsistent. If $ABCD$ is a convex quadrilateral with $AB=BC=AC$, then the angle $\angle BDC$ will be restricted to the range $60^{\circ} \leq \angle BDC \leq 90^{\circ}$. Mar 2 '14 at 8:00
• @DavidH Can you elaborate? I did not get it [added the diagram]. Mar 2 '14 at 8:03
• OK, the diagram you added clears up the problem. In a quadrilateral $ABCD$, the vertex $D$ is connected to $C$ and $A$ and is opposite vertex $B$. But in your diagram, you changed the order of the vertices to $ABDC$. Mar 2 '14 at 8:13
• @DavidH Ooh, big mistake! Do you have any ideas? It is an apparently simple question, but has weathered all my continuous attempts! Mar 2 '14 at 8:18
• Draw the circumcircle of $\triangle BCD$, and let its center by $K$. Point $K$ necessarily lies on the perpendicular bisector of $\overline{BC}$; by the Inscribed Angle Theorem, $K$ is such that $\angle BKC = 360^\circ - 2\angle BDC = 60^\circ$. Consequently, $K=A$, and the result follows.
– Blue
Mar 2 '14 at 8:32

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.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Let's $\ds{\verts{\vec{\rm AB}} = \verts{\vec{\rm BC}} = \verts{\vec{\rm AC}} \equiv \ell}$
$$\ell\ \overbrace{\sin\pars{60^{o}}}^{\ds{\root{3}/2}} + {\ell/2 \over \underbrace{\tan\pars{75^{o}}}_{2 + \root{3}}} = \pars{{\root{3} \over 2} + \half\,{\root{3} - 2 \over 3 - 4}}\,\ell ={\root{3} - \pars{\root{3} - 2} \over 2}\,\ell = \color{#00f}{\Large\ell}$$
• First term is he height of the equilater triangle. The second one is the small piece which bisects the 150$^{o}$ angle. The sum is the $\vec{AD}$ length $\left\vert\vec{AD}\right\vert$. Mar 2 '14 at 9:22