Let $ABC$ a triangle with right angle at $C$. Let $D\in AB$ such that $CD\perp AB$. Construct the triangle given $AC$ and $BD$.
Using the relation $AC^2=AD(AD+BD)$ I tried to construct a segment of length $AD$ using some similarities, but I did not manage it. Also, I tried rotations, but again without an outcome.
So, any suggestions, please?
Introduce $r=AC$, $CD=h$, $AD=p$, $DB=q$. You know $r$ and $q$.
From triangle ADC:
$$r^2-p^2=h^2=pq$$
$$p^2+pq-r^2=0$$
$$p=\frac{-q+\sqrt{q^2+4r^2}}{2}$$
$$p=\frac{\sqrt{q^2+(2r)^2}-q}{2}\tag{1}$$
Construction follows from (1):
Construct right angled triangle with legs $q$ and $2r$. Its hypothenuse has length $\sqrt{q^2+(2r)^2}$. Subtract $q$ from it and find a half of the resulting segment. You have constructed $p$.
Now that you have $p,q,r$ you should be able to complete the construction easily.