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! !enter image description here **EDIT** : I don't want to use vectors, only standard geometry and trigonometry.

Converting comment to answer with a few more details, as requested.

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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.