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?
