Quadrilateral ABCD satisfies $\overline{2AB}=\overline{AC}$, $\overline{BC}=\overline{\sqrt{3}}$, $\overline{BD}=\overline{DC}$ and $Quadrilateral ABCD is inscribed in a circle satisfies $\overline{2AB}=\overline{AC}$, $\overline{BC}=\overline{\sqrt{3}}$, $\overline{BD}=\overline{DC}$ and $<BAC=60$
I've never been good at Euclidean geometry questions like this... Really, what strategies could I employ to begin an analysis of the situation? I'm working through a text on Euclidean geometry and this is a question. The specific questions being asked about this scenario are the following:
1). The radius of the circumscribed circle
2). $\overline{AC}$
3). $<BDC$
4). The area of $\Delta BDC$
5). $\overrightarrow{CA} \cdot \overrightarrow{DC}$
Now, The fact that $\overline{BD}=\overline{DC}$ struck me as odd. The length of a diagonal is equal to the length of one of the sides? I want to say that this implies that the quadrilateral we are dealing with must not be convex.
Since $<BAC=60$, we can use the law of cosines to deduce that $\overline{AC}=2$ and so $\overline{AB}=1$.
My analysis sort of hits a road block here. I have not used the fact that the quadrilateral is inscribed in a circle. For a quadrilateral inscribed in a circle, it is well known that:

*

*The product of the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the product of its two pairs of opposite sides.


*

*The opposite angles of  quadrilateral inscribed in a circle are supplementary. i.e., the sum of the opposite angles is equal to 180˚.

However, I have not been able to make use of these facts.
Can anyone help me out here??  Thanks in advance!

*

*First note that since $<BAC = 60$ and $2AB = AC$ it tells us $\Delta ABC$ is congruent to the general $30, 60, 90$ triangle (by $SAS$). This tells us that $<ABC = 90$.
 Using the fact that the hypotenuse of an inscribed right triangle is a diameter then we have that $r = \frac{AC}{2}$ but as you stated $AC = 2$ so $r = 1$


*Correctly by using laws of cosines $AC = 2$


*Note that we know $<BAC = 60$ and $<BDC$ lies on the same arc as $<BAC$ (arc $BC$) then we know these angle equal so $<BDC = 60$


*Now we know that $<BDC = 60$ we can see that $\Delta BDC$ is an congruent to the general equilateral triangle (by SAS since $BD = DC$). This now tells us $BD = DC = BC = \sqrt{3}$ so the area of $\Delta BDC = \frac{1}{2}\sqrt{3}\sqrt{3}sin(60) = \frac{3\sqrt{3}}{4}$


*We have already worked out that $|CA| = 2$ and $|DC| = \sqrt{3}$. Then since $<ABC = 90$ and $<DBC = 60$ we have that $<ABD = 30$ and therefore since angles in the triangle $\Delta ABX$ (where $X$ the intersection of lines $AC$ and $BD$) is add to $180$, lines BD and AC meet at $90$ degrees. Hence $\Delta BCX$ is congruent to $\Delta DCX$ (by $AAS$) which tells us that the angle between $CA$ and $DC$ is equal to $<ACB = 30$. Hence the dot product of $CA$ and $DC$ is $-2\sqrt{3}\cos{30} = -3$
If you want a short extension then show that quadrilateral $ABCD$ is a kite
