Find locus of intersection of two lines with parameters $\require{cancel}$
I have a triangle

where $$DE || BC, A(a,0), B(0,b), b>0, a>0$$
I need to find the locus of the intersection between $BE$ and $CD$ (in terms of $a$ and $b$).
They didn't tell me more about this parallel line that cuts AB y and AC, therefore I thought I would need to create some variables such that
$$E (e,0) \quad \text{and} \quad D(e,d)$$
From here I get from Tales's theorem that
$$\frac{DE}{BC} = \frac{AE}{AC} \Rightarrow \frac db = \frac{a-e}{a} \Rightarrow d = \frac{b(a-e)}{a}$$
$$\sqrt{a^2-2ae+e^2+d^2} = AD \quad \text{Pythagoras theorem}$$
$$AB = \sqrt{a^2+b^2}$$
$$\tag{$()^2$}\frac{\sqrt{a^2-2ae+e^2+d^2}}{\sqrt{a^2+b^2}} = \frac db$$
$$\frac{a^2-2ae+e^2+d^2}{a^2+b^2} = \frac {d^2}{b^2}$$
$$\frac{a^2-2ae+e^2 + \frac{b^2\left(a^2-2ae+e^2\right)}{a^2}{}}{a^2+b^2} = \frac{\frac{\cancel{b^2}\left(a^2-2ae+e^2\right)}{a^2}}{\frac {\cancel{b^2}}{1}}$$
At the end I would get something like
$$(a^2-2ae+e^2)(a^2+b^2) = a^2(a^2-2ae+e^2) + b^2(a^2-2ae+e^2)$$
Which will not help. I'm doing something wrong. I wanted to find $e$ and $d$ in terms of $a$ and $b$ so I can find the locus I need.
 A: My idea is to try to express the coordinates of $F = (x,y)$, the intersection of $BE$ and $CD$, with $a$, $b$ and $E = (e,0)$. Let $G$ be the point on x-axis such that $FG\perp CE$. $$\frac{FG}{DE}+\frac{FG}{BC} = 1\implies y = FG = \frac{DE\cdot BC}{DE+BC}$$ 
With $\frac{DE}{BC} = \frac{a-e}{a}\implies DE = \frac{ab-be}{a}$, we have $y=\frac{ab-be}{2a-e}$, which is equivalent to:
$$e = \frac{ab-2ay}{b-y}$$
Further, we can derive a result for $x$:
$$\frac{EG}{CE} = \frac{FG}{BC}\implies \frac{e-x}{e} = \frac{a-e}{2a-e}\implies x=e-\frac{ae-e^2}{2a-e}$$
and we can put the above two equations together (by plugging $e$ expressed as $a$, $b$, $y$ into the above equation) to get the locus
$$x=a-\frac{2a}{b}\cdot y$$
or equivalently, $$y=\frac{b}{2}-\frac{b}{2a}\cdot x$$
A: Parameterize the point $D$ as $(1-t)A+tB=((1-t),bt)$, $t\in[0,1]$. An equation for the line $\overline{CD}$ is then $btx-(1-t)ay=0$. The point $E$ is $((1-t)a,0)$, so an equation for $\overline{BE}$ is $bx+(1-t)ay=(1-t)ab$ and the intersection of these two lines is $x={1-t\over1+t}a$, $y={t\over1+t}b$. From here, it’s a simple matter to eliminate $t$: $x-a=-{2at\over1+t}$, so ${y\over(x-a)}=-{b\over2a}$, therefore the locus of intersections is the line $y=-\frac b{2a}(x-a)$.
