I'm reading Martin Liebek's book, and he gives a construction of $\sqrt{n}$ as the side length of a triangle, but leaves the verification as an exercise. It's a bit difficult to describe the picture in full without drawing it, but the construction is:
- We have a circle with diameter $AB$, with $D$ being a point between $A$ and $B$.
- The point $C$ is located on the circle directly above the point $D$.
- $D$ is chosen so that $AD = n$ and $DB = 1$.
- The triangles $ACD$, $BCD$, and $ABC$ are right triangles.
The goal is to solve for side $CD$, which he claims is equal to $\sqrt{n}$. From the picture, it isn't immediately clear to me that $ACD$ is a right triangle, so I assume this is an assumption baked into the problem, and there isn't a way to "prove" this is the case. I'll try to apply the Pythagorean theorem to each triangle:
\begin{align*} (AD)^2 + (CD)^2 &= (AC)^2 \\ (DB)^2 + (CD)^2 &= (CB)^2 \\ (AC)^2 + (CB)^2 &= (AB)^2 \end{align*} This is a system of three equations in three unknowns, so it should be solvable, but I'm struggling with the verification. To make things a bit simpler, I'll replace $AD$ with $n$, $DB$ with $1$, $AB$ with $n+1$, and define \begin{align*} a &= (CD)^2 \\ b &= (AC)^2 \\ c &= (CB)^2, \end{align*} so the system becomes \begin{align*} n^2 + a^2 &= b^2 \\ 1 + a^2 &= c^2 \\ b^2 + c^2 &= (n+1)^2. \end{align*} We substitute $n^2 + a^2$ for $b^2$ in equation (3): $$ n^2 + a^2 + c^2 = (n+1)^2 = n^2 + 2n + 1, $$ so, cancelling $n^2$, we get $$ a^2 + c^2 = 2n + 1. $$ We substitute $1 + a^2$ for $c^2$, and cancel a $1$ to get $$ 2a^2 = 2n, $$ so $a^2 = n$, so $a = \sqrt{n}$. That is, $CD = \sqrt{n}$.
How does this proof look?
You did a nice proof by applying Pythagorean Theorem. And an alternative solution by using similarity between triangles is
$\angle ACB=90^{\circ}$ (angle in semicircle)
$\angle CDA=\angle CDB=90^{\circ}$
$\angle A=\angle BCD$
$\angle ACD=\angle B$
$\triangle CAD \sim \triangle BCD$ (AAA Similarity Theorem)
$\frac{AD}{CD}=\frac{CD}{BD}$
$\frac{n}{CD}=\frac{CD}{1}$
$CD=\sqrt n$