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!
 A: *

*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
A: Let $x = \overline{AB}$ , then $\overline{AC} = 2 x$, and we are given that $\overline{BC} = \sqrt{3} $ and that $\angle BAC = 60^\circ$
Applying the law of cosines to $\triangle ABC$, we get
$ (\sqrt{3})^2 = x^2 + (2 x)^2 - 2 x (2 x) (\frac{1}{2}) $
Hence,
$ 3 = x^2 + 4 x^2 - 2 x^2 = 3 x^2 $
From which, $x = 1$.  Therefore, $\overline{AB} = 1 $ and $\overline{AC} = 2 $
Now using coordinate geometry, we can set vertex $A = (0,0)$ and $B = (1, 0)$, then $C = (1, \sqrt{3}) $ (Assuming $C$ is in the upper half plane).
Next, we find the circle on which $A,B,C$ lies.  The center is the intersection of the perpendicular bisector of $AB$, which is the line $x = \dfrac{1}{2}$, and the perpendicular bisector of $BC$, which is the line $ y = \dfrac{\sqrt{3}}{2} $
Hence the center of the circumcircle of $\triangle$ ABC is $O = (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} ) $
The radius of this circumcircle is $R = \overline{OA} = 1 $
Since point $D$ lies on the cicrcumcircle and on perpendicular bisector of $BC$ (because it is equidistant from $B$ and $C$), then we $D$ can be
$D_1 = O + (-R, 0) = (-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} ) \hspace{25pt} (1)$
or
$D_2 = O + (R, 0) = ( \dfrac{3}{2} , \dfrac{\sqrt{3}}{2} ) \hspace{25pt} (2)$
However, from the cyclic order of the vertices, $D_2$ is rejected.  Hence $D = D_1 $.
Now we can compute the area.
The area of $ABCD = [ABC] + [ACD_1] = \frac{1}{2} \left( \sqrt{3} + \dfrac{\sqrt{3}}{2} - \sqrt{3} (- \dfrac{1}{2} ) \right) = \sqrt{3} $
Also, we have $\vec{CA} = A - C = (-1, -\sqrt{3} ) $, and $\vec{DC} = C - D = ( \dfrac{3}{2} , \dfrac{\sqrt{3}}{2} ) $
So $\vec{CA} \cdot \vec{DC} = - \dfrac{3}{2} - \dfrac{3}{2} = -3 $
