This is quite an easy question, but it's been troubling me and I can't manage to work it out. I've been reading the book A Concise Introduction to Pure Mathematics (M. Liebeck), so I'll quote the relevant passage here:
The following slightly more complicated geometrical argument than that given in Proposition 2.2 shows the existence of the real number $\sqrt{n}$ for any positive integer $n$. As in the figure below, draw a circle with diameter $AB$, with a point $D$ marked so that $AD = n$, $DB = 1$. We leave it to the reader to use Pythagoras in the right-angled triangles $ACD$, $BCD$ and $ABC$ to show that the length $CD$ has square equal to $n$, and hence $CD = \sqrt{n}$
I don't have a scanner at hand, but luckily enough, this particular figure is shown on the cover of the book (bottom-right corner), so I'll attach an image of the cover (the only difference is that point $D$ is not labelled on the cover for some reason, but, as you can probably work out, it's the unlabelled point on the hypotenuse of the largest triangle where the two smaller triangles meet). The problem I have is that everything I know about the Pythagorean theorem requires that you already know two sides of a triangle in order to find the third (i.e. $a^2 + b^2 = c^2$), and yet only one is given for each of the triangles. I therefore presume that there are inferences that can be made involving the three triangles, but I don't understand what, as I'm not a professional mathematician or even student, I'm just teaching myself out of interest, so am very grateful for any help. Thank you.
Suppose $CD=x$. Then $x^2+n^2 = AC^2$ and $1+x^2 = BC^2$. Adding these two equations and using Pythagoras again gives $$2x^2 + n^2 + 1 = AC^2 + BC^2 = AB^2 = (n+1)^2 = n^2 + 2n + 1,$$ so that $2x^2 = 2n$ and $x^2 = n$.