Locus of a moving point, when constraints on an angle and length are given

Question

$APQ$ is a variable triangle; $A$ is fixed, $P$ moves on a fixed line $CD$; if $AP$ meets a fixed line parallel to $CD$ at $R$, and if $PQ=AR$ and if the angle $APQ$ is constant, prove that the locus of $Q$ is a straight line.

I have attempted some angle chasing. I wish to show that for two such points $Q_1$ and $Q_2$, the angle made by $Q_1Q_2$ with $CD$ is independent of the angle $P_1-A-P_2$, and thus where $Q_1$ and $Q_2$ are chosen. The only concrete progress I have made as of now is noticing that angle $PQR$ is constant.

Answer 1

Note that by similarity, $AR = k\cdot AP$ for some fixed $k \in \mathbb R$. So we have $PQ = k \cdot AP$, which means that $\frac{PQ}{AP} = k$. $\angle APQ$ constant tells us that $\triangle APQ$ is always similar, i.e. taking $A$ as the origin, and $p, q$ as the locations of $P$ and $Q$ in complex coordinates, $q = wp$ for some fixed complex number $w$. This is a composition of a rotation and homothety, and thus sends the line that $P$ is on to a line that $Q$ is always on.

Answer 2

Here is an analytical geometry proof.

WLOG, one can take coordinate axes such that $A(0;0)$, and line $CD$ is the horizontal line with equation $y=1$ ; consequently, due to alignment of $R$ with $O$ and $P$, $R$ has coordinates $\binom{kx}{k}$ (therefore, the locus of $R$ is horizontal straight line with equation $y=k$, $k$ constant).

The unit vector of axis $\vec{AP}$ is $\vec{U}=\dfrac{1}{\sqrt{x^2+1}}\binom{x}{1}$

Therefore, $Q$ is such that, if we denote by $\alpha$ the fixed angle, and by $L$ the length of $AR$ with $L=\sqrt{(kx)^2+k^2}=k\sqrt{x^2+1}$ :

$$\vec{AQ}=\vec{AP}+\vec{PQ}=\binom{x}{1}+L.Rot_{\alpha}\vec{U}$$

$$=\binom{x}{1}+k\sqrt{x^2+1}\begin{pmatrix}\cos \alpha&-\sin \alpha\\ \sin \alpha&\ \cos \alpha \end{pmatrix}\dfrac{1}{\sqrt{x^2+1}}\binom{x}{1}$$

$$=\binom{x}{1}+k\begin{pmatrix}\cos \alpha&-\sin \alpha\\ \sin \alpha&\ \cos \alpha \end{pmatrix}\binom{x}{1}$$

$$=\begin{pmatrix}1+k\cos \alpha&-k\sin \alpha\\ k\sin \alpha&\ 1+k\cos \alpha \end{pmatrix}\binom{x}{1}$$

which is a linear transform of a straight line, thus a straight line itself.