# Constructing $\sqrt{a}$ for a constructable $0\leq a\in\mathbb{R}$ - Compass and straightedge constructions [duplicate]

Possible Duplicate:
Compass-and-straightedge construction of the square root of a given line?

I wish to understand how to construct $\sqrt{a}$ for a constructable $0\leq a\in\mathbb{R}$ , the book Abstract Algebra (by David Steven Dummit, Richard M. Foote) offers (in pg. 532) the following:

construct the circle with diameter $1+a$ (looks like a straight line with the point $a$ somewhere on the line and $1+a$ at the right end of the line) and erect the perpendicular to the diameter from the point $a$ (the point with distance $a$ from the leftmost point on the line). Then $\sqrt{a}$ is the length of the perpendicular.

My question is why the length of the perpendicular is $\sqrt{a}$ ? (I'm guessing that there's a theorem in geometry that I don't know about that might help...)

Help is appreciated!

-

## marked as duplicate by Alex Becker♦, Ｊ. Ｍ., Leonid Kovalev, Asaf Karagila, t.b.Aug 16 '12 at 12:45

See in particular this answer to the question @Tim found. –  Henning Makholm Jul 8 '12 at 21:47

This is simple geometry (Pythagorean theorem):

Take the circle with diameter $a+1$. then the point $a$ is at distance $\frac{a-1}{2}$ from the circle's center. the radius is $\frac{a+1}{2}$. So the perpendicular satisfies $(\frac{a-1}{2})^2+x^2=(\frac{a+1}{2})^2$, thus $x=\sqrt{a}$.

-

You can show it without the Pythagorean theorem, in a more geometric but longer proof.

Let $[BA]$ be the segment of length $a$, $[AC]$ its prolongation of length 1, and $H$ in on the circle of diameter $[BC]$ such that $AH\perp BC$.

1. Thee triangle $CHB$ is rectangle in $H$ :

Let $O$ be the centre of the circle. Since the triangles $COH$ and $BOH$ are isosceles, one $\angle OCH=\angle OHC$ and $\angle OBH=\angle OHB$. Therefore, $\angle CHB=\angle BCH + \angle CBH$. Since the sum of the angles of $CBH$ is $\pi$, one has $CBH=\pi/2$

2. The triangles $CAH$ and $AHB$ are similar :

Each of them has two angles in common with the $CHB$, so they are both similar to it.

3. End of the proof :

Since $CAH$ and $AHB$ are similar, $\frac{CA}{AH}=\frac{AH}{AB}$, or equivalently $AH^2=CA\cdot AB=1\cdot a$.

Note that a slight variation on this reasoning gives another construction. If you consider the similar triangles $BHC$ and $HAC$, you have $\frac{HC}{BC}=\frac{AC}{HC}$, or $HC=\sqrt{BC\cdot AC}$. $[HC]$'s length is the square root of the circle's diameter.

-