# Ruler and compass construction [duplicate]

This question already has an answer here:

Given the three line segments below, of lengths a, b and 1, respectively:

construct the following length using a compass and ruler: $$\frac{1}{\sqrt{b+\sqrt{a}}} \ \ \text{and} \ \ \ \sqrt[4]{a}$$

Make sure to draw the appropriate diagram(s) and describe your process in words. We are also to use the following axioms and state where they are used:

1. Any two points can be connected by a line segment,
2. Any line segment can be extended to a line,
3. Any point and a line segment define a circle,
4. Points are born as intersection of lines, circles and lines and circles

Can someone please guide me or show me as to how to construct this? I know if we draw a triangle whose base(let's suppose this is $a+1$) is the diameter of a semi-circle, then the line perpendicular to this base leading to the top of the semi-circle will divide the trianlge into two smaller triangles with the bases resulting in $a$ and $1$. I don't know how to end up with $\sqrt{a}$ from there. But with it, the process can be repeated to end up with $\sqrt[4]{a}$. Can someone explain or show me? I will then be able to tackle a whole lot of other questions.

## marked as duplicate by C. Falcon, J. M. is a poor mathematician, Juniven, Community♦Apr 8 '17 at 2:38

• just to clarify: for the first length to be construced, it is 1 / (sqrt(b + sqrt(a))? – Cursed Apr 7 '17 at 20:53
• @Cursed1701 Yes, exactly! – Yahya Farooq Apr 7 '17 at 20:54
• @YahyaFarooq I suggest you edit the mathjax then, as thats not quite what it sais – Cursed Apr 7 '17 at 20:54
• The triangle you desccribe will have altitude h. a/h = h/1 so h^2 = a^2. – fleablood Apr 7 '17 at 20:56

It is all similar right triangles, along with the theorem that, when a triangle has all three vertices on a circle and two of them on a diameter, then it is a right triangle.

The triangles you described will have AB=1. BC=a AC=a+1. and BD a perpendicular altitude of unknown height. Triangle ABD is similar to triangle DBC.

So $\frac {AB}{BD} = \frac {DB}{BC}$

so $\frac {1}{BD} = \frac {BD}{a}$

so $a = BD^2$

so $BD = \sqrt{a}$.

The height is no longer unknown.