# 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

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.


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

• Which two lengths you have added in the first step? Mar 2 '14 at 9:09
• 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
• Wait a minute, how do you prove that those two add up to the diagonal. Mar 3 '14 at 7:08