Given a triangle,a similar triangle $ABC$ is drawn where $A$ is a fixed point and $B$ is any point on a given line.Find the locus of $C$. Given a triangle,a similar triangle $ABC$ is drawn where $A$ is a fixed point and $B$ is any point on a given line.Find the locus of $C.$
I thought over it but there is no concrete method visible to find out the locus of C.
 A: Assume that $ABC$ is a triangle and $B$ belongs to some line $l$.
Let $D$ be the intersection $\neq B$ between $l$ and the circumcircle of $ABC$.
Let $C'$ be some point on the line $DC$, and $B'\in l$ a point such that $\widehat{B'AC'}=\widehat{BAC}$.
By angle chasing we have that $B'A C'$ is similar to $BAC$, since both $BACD$ and $B'AC'D$ are cyclic quadrilaterals. It follows that the wanted locus is a line.

A: Let us consider in the complex plane.
We may suppose that $A(\alpha)$ is at $\alpha=i$ and that $B(\beta)$ is on the real axis ($\beta=p$ where $p\in\mathbb R$).
Also, if we set $q=\frac{A'B'}{C'A'},\theta=\angle{C'A'B'}$, then for $C(\gamma)$, we have
$$\gamma-\alpha=q(\cos\theta+i\sin\theta)(\beta-\alpha)$$
where we suppose that $\triangle{A'B'C'}$ is given and that $\triangle{ABC}$ is similar to $\triangle{A'B'C'}$.
Then,
$$\begin{align}\gamma &=i+q(\cos\theta+i\sin\theta)(p-i)\\&=i+q(p\cos\theta-i\cos\theta+p\sin\theta i+\sin\theta)\\&=(pq\cos\theta+q\sin\theta)+i(1-q\cos\theta+pq\sin\theta)\end{align}$$
So, if we set $\gamma=x+yi$ where $x,y\in\mathbb R$, then we have
$$x=pq\cos\theta+q\sin\theta,\quad y=1-q\cos\theta+pq\sin\theta$$
Eliminating $p$ from these gives
$$(x-q\sin\theta)q\sin\theta=q\cos\theta(y-1+q\cos\theta),$$
i.e.
$$y-(1-q\cos\theta)=\tan\theta\ (x-q\sin\theta).$$
By the way, we can replace $\theta$ with $-\theta$.
Hence, the answer is that the locus of $C$ is two lines :
$$y-(1-q\cos\theta)=\tan\theta\ (x-q\sin\theta),$$
$$y-(1-q\cos(-\theta))=\tan(-\theta)\ (x-q\sin(-\theta)).$$
A: We may assume $A=0\in{\mathbb C}$, $B=b+t\, i$ with $b>0$ and $t\in{\mathbb R}$. On the other hand, let $\triangle(A_*B_*C_*)\subset {\mathbb C}$ be the given sample triangle, and define $c\in{\mathbb C}\setminus{\mathbb R}$ by
$$C_*-A_*= c(B_*-A_*)\ .$$
The locus in question is then parametrized by
$$t\mapsto C(t):=c(b+t\,i)\qquad(-\infty<t<\infty)\ .$$
It is a line through the point $C_0:=c \,b$, and orthogonal to $c$.
