Coordinate Geometry Oblique Coordinates Problem This is a elementary geometry problem which I have tried to  solve using coordinate geometry but it is resulting in an impossible and impractical  result. Maybe I have some misconceptions with oblique coordinates. So please help me with my misconceptions.

Suppose $\triangle ABC$ is a triangle in which $\angle B= 2 \angle C$. $D$ is a point on $BC$ such that $AD$ bisects $\angle BAC$ and $AB=CD$. Prove that $\angle BAC= 72^\circ$.

My solution
Let us construct the coordinate system in $\triangle BAC$ where the $X$-axis is along the side $BC$ and the $Y$ axis is along the side $AD$ with angle of origin=$\angle G$. Consider $D$ as the origin with coordinates $(0,0)$. Let the coordinates  of $C$ be $(a,0)$ and that of $A$ be $(0,c)$. Let the coordinates   of $B \equiv (b,0)$.
Then the equation of straight line $AC=\frac x a + \frac y c=1$ (by intercept form). That is, $cx + ay= ac$, i.e., $ cx+ay-ac=0$. Equation of straight line $\frac x{-b} + \frac y c =1$, i.e., $cx -by= -bc$, i.e., $cx-by+bc=0$.
As $AD$ bisects angle $\angle BAC$, the angle between $cx-by+bc= 0$ and $x=0$ = to the angle between $cx+ ay-ac=0$ and $x=0$. So their tangents are equal, i.e., $\frac{b \sin g}{c+b \cos g}= \frac{a \sin g}{c-a \sin g}$.
Cancelling $\sin g$ from both sides  we have $- bc + ab \cos g =ac + ab \cos g $. Cancelling $ab \cos g$ from both sides we have $ac+ bc=0$  ie $a+b=0$
ie $a=-b$ This implies $DC= BD$. But by angle bisector theorem we have $\frac{AB}{AC}= \frac{BD}{DC}$ as $BD = DC$ we have $AB = AC$. Thus the $\triangle ABC$ is isosceles which is contradictory by the question.
How is this possible?


There are two difficulties here, as the post stands.  One is that it appears as though there was an earlier version of the solution posted, which was then revised, leaving it in a form that is now internally inconsistent.  Unless, for some reason, two coordinate axes are being used which are not perpendicular to one another (which would require significant modification to the trigonometric ratios shown), it cannot be the case that both point $ \ D \ $ is placed at the origin of the coordinate system and that point $ \ A \ $ lies on the $ y-$ axis.  This is evidently the cause of the false conclusion that $ \ DC = BD \ . $  (There are isoceles triangles in this diagram: $ \ ABC \ $ is one of them, but not with the congruence of sides suggested.)
The other problem is with taking a purely analytical-geometric approach here.  Since we wish to show that $ \angle BAC \ $ has a measure of 72º , we will eventually run up against the issue of whether we will recognize the exact value of a trigonometric ratio for that angle when we encounter it.
Application of trigonometry is a useful approach; I will lay out one particular route here.

I have marked the given information for the triangle in blue; we will derive the angles marked in red.  The angle $ \ \angle BAC \ $ has measure $ \ 180º - 3 \theta \ , $ so we have $ \ m(\angle BAD) = m(\angle CAD) $ $ = 90º - \frac{3}{2} \theta \ . $  It follows by one or another line of argument that $ \ m(\angle ADC)  = 90º + \frac{1}{2} \theta \  $ and $ \ m(\angle ADB)  = 90º - \frac{1}{2} \theta \ .  $
The specification that $ \ AB = CD \ $ (the lengths of which we shall call $ \ c \ $ ) is crucial to resolving this triangle.  We may use the Law of Sines to write for triangle $ \ ABD \ , $ 
$$ \frac{\sin (90º - \frac{1}{2} \theta)}{c} \ = \ \frac{\sin 2 \theta}{d} \ ,   $$
and for triangle $ \ ACD \ , $
$$ \frac{\sin (90º - \frac{3}{2} \theta)}{c} \ = \ \frac{\sin  \theta}{d} \ .   $$
From these ratios and the use of the "co-relations", we obtain $ \ c \ \sin \theta \ = \ d \ \cos \frac{3}{2} \theta \ $ and $ \ c \ \sin 2 \theta \ = \ d \ \cos \frac{1}{2} \theta \ . $  Using the "double-angle formula" for sine and substituting the first equation into the second produces
$$ \ c \ \cdot 2 \ \sin  \theta \ \cos \theta \ = \ d \ \cos \frac{1}{2} \theta \ \ \Rightarrow \ \ \ 2d \ \cos \frac{3}{2} \theta \ \cos \theta \ = \ d \ \cos \frac{1}{2} \theta  \ , $$ 
from which we may cancel the factor $ \ d \ $ .  The appropriate "product-to-sum formula" then leads us to
$$ 2 \ \left( \frac{1}{2} \ \left[ \ \cos (\frac{3}{2} \theta \ - \ \theta) \ + \ \cos (\frac{3}{2} \theta \ + \ \theta) \ \right] \ \right) \ = \  \cos \frac{1}{2} \theta $$
$$  \cos (\frac{1}{2} \theta ) \ + \ \cos (\frac{5}{2} \theta ) \  = \  \cos \frac{1}{2} \theta \ \ \Rightarrow \ \ \cos (\frac{5}{2} \theta ) \  = \  0 \ \ . $$
We may restrict ourselves to the first quadrant for a solution, obtaining $ \ \frac{5}{2} \theta  \  = \  90º \ \Rightarrow \ \theta \ = \ 36º . $  At last, we conclude that 
$$ \ m(\angle ACB) \ = \ 36º \ \ , \ \ m(\angle ABC) \ = \ 72º \ \ , \ \text{and} \ \ m(\angle BAC) \ = \ 72º \ .  $$
$$ \ \ $$
Now for the more interesting aspect of this figure: with a little more work, we find that $ \ \ m(\angle BAD) \ = \ 36º \ \  \text{and} \ \ m(\angle ADB) \ = \ 72º \ .  $  So, in fact, $ \ ACB \ $ and $ \ BAD \ $ are similar isosceles triangles.  Therefore, we have a proportionality which tells us that
$$ \ \frac{c+x}{c} \ = \ \frac{c}{x} \ \ \Rightarrow \ \ x^2 \ + \ cx \ = \ c^2 \ ,  $$
which can be solved alternately to find $ \ c \ = \ x \ \left(\frac{1 + \sqrt{5}}{2} \right) \ = \ \phi \ $ or $ \ x \ = \ c \ \left(\frac{ \sqrt{5} - 1}{2} \right) \ = \ \frac{1}{\phi} \ . $  Sides $ \ BC \ ( \text{or} \ AC ) \ $ and $ \ AB \ $ and sides $ \ AB \ ( \text{or} \ AD ) \ $ and $ \ BD \ $ stand in proportions of the Golden Ratio!
Our triangle diagram is actually a portion of the so-called "Golden Pentagram", sacred to the Pythagoreans.  (When the dominant culture of the region changed in subsequent centuries, the connotations associated with this figure followed the (somewhat simplistic) path, Greek $ \ \rightarrow \ $ pagan $ \ \rightarrow \ $ infidel $ \ \rightarrow \ $ evil.)  This figure is rife with mathematical interest; see, for instance, the recent Posamentier and Lehmann, The Glorious Golden Ratio (2012), pp. 137-60.

