I want to prove the following theorem:
Let $g$ be a line tangent to the circle $k$, $P$ an arbitrary point and $h$ the parallel to $g$ tangent to $k$ (where $h\neq g$). Show that point $Q, R$ exist on $g$ such that $k$ is the incircle of $\Delta PQR$ iff $P$ is located in the other half-plane than $g$ with respect to $h$ (and is not on $h$).
Can you verify my attempt for a proof of that Theorem? It goes as follows:
First, let $Q, R$ be on $g$ such that $k$ is the incircle of $\Delta PQR$ and let $p$ be the parallel to $g$ through $P$. Then $p$ is located entirely outside of $PQR$, $k$ is located inside. Thus, the two have no point in common. If one now translates $p$ along a vector perpendicular to $g$ in the direction of $g$, $p$ has to pass the boundary of $k$ after a positive amount of time $t>0$ since $k$ is located between $p$ and $g$ and translation is continuous. When $p$ and the boundary of $k$ have at least one point in common for the first time, $p$ has to be tangential to $k$ since it is parallel to the Tangent $g$. This has to be the Tangent $h$ and because of the Translation, $P$ is located in the other half-plane than $g$ with respect to $h$ (and is not on $h$ because of $t>0$).
Now let $p$ be located in the other half-plane than $g$ with respect to $h$ (and $P\notin h$) and let $p$ be the parallel to $g$ through $P$. $P$ subdivides $p$ into two rays $p_1$ and $p_2$. Because of $p\neq h$ and the location of $P$, $p_1$ and $k$ have no point in common. If one now rotates $p_1$ by $180^{\circ}$ around $P$ such that at least once during the rotation, it has a point in common with $g$, then $p_1$ ends up lying on $p_2$ again having no point in common with $k$. During the rotation, because of the continuity of the rotation and the location of $P$, it must have passed over the boundary of $k$ twice, i.e. it was twice tangential to $k$. Now if one chooses the intersection of these two tangents with $g$ as $Q$ and $R$, $k$ is the incircle of $PQR$.
Is this correct/rigorous? I appreciate any form of advice.
I think your proof for the first part is not rigorous.
Let $$\begin{align}P_1&:\text{The translation always works} \\\\P_2&:\text{$P$ is located in the other half-plane than $g$ with respect to $h$ and is not on $h$}\end{align}$$
You are saying that "$P_1\implies P_2$ is true, so $P_2$ is true". Well, you are correct that $P_1\implies P_2$ is true. However, you have not proven that $P_1$ itself is true. So, as a result, you have not proven that $P_2$ is true. In short, you need to write a proof that the translation always works even if it looks very obvious.
I would prove the first part in the following way :
Let $A,B,C$ be the tangent point to $k$ of $PQ,QR,RP$ respectively. Also, set $QA=QB=a,PA=PC=b,RB=RC=c$. Let $d$ be the distance from $P$ to $g$, and let $r$ be the radius of $k$.
Representing the area of $\triangle{PQR}$ in two ways, we get $$\frac 12(a+c)d=\frac r2(a+b)+\frac r2(b+c)+\frac r2(c+a)$$ from which $$d=2r+\frac{2br}{a+c}\gt 2r$$ follows.$\quad\square$