Let $P \in H^2,$ and $L$ be a complete geodesic in $H^2.$
Prove that there is a unique complete geodesic $L'$ through $P$ and orthogonal to $L$ at some point $Q \in L.$ Moreover, the line segment from $P$ to $Q$ minimizes the distance from $P$ to any point on $L.$
I am not confident in my proof:
I do not understand isometries well and and may have used them incorrectly.
Am I on the right track?
Solution:
(1) Consider the case where $L$ is a vertical half-line $$L = \{(x_0,y) \in R^2; y > 0\}.$$ Easily, one can construct a complete geodesic of the hyperbolic plane by centering a half circle at $x_0$ and having it contain the point $P.$ The geodesic $L'$ intersects $L$ orthogonally.
Now, I consider any complete geodesic $L$ in the hyperbolic plane. I state that I can transform this geodesic into the vertical half-line via some compositions of isometries $\phi.$ Since there exists a geodesic $L'$ that contains $P$ and intersects $\phi(L)$ orthogonally at some point $Q',$ and since isometries preserve distances and angles, and since the inverses of isometries are isometries, it follows that $\phi^{-1}(L')$ intersects $L$ orthogonally at point $Q = \phi^{-1}(Q')$ while still containing the point $P.$
I use a similar argument to attempt the second question.
(2) By using some composition of isometries $\phi,$ I can convert $L'$ into the vertical half line. I know that the distance between $\phi(P)$ and $\phi(Q)$ is smaller than the distance between $\phi(P)$ and any other point on the geodesic $\phi(L').$ So, the same is also true for $P$ and $Q.$