Divide a line segment in the ratio $\sqrt{2}:\sqrt{3}.$ "Divide a line segment in the ratio $\sqrt{2}:\sqrt{3}.$"
I have got this problem in a book, but I have no idea how to solve it.
Any help will be appreciated.  
 A: I assume you need to construct said division.
Let $AB$ be the line segment to be divided. Construct the semilines $r$ and $s$ perpendicular to $AB$ through $A$ and $B$, respectively such that they are both in the same "side" of $AB$.
Draw a circle of radius $AB$ from $A$. It intersects $r$ in $C$. Draw a circle of radius $BC$ from $B$. It intersects $s$ in $D$. Now $BD = \sqrt{2}AB$. Draw a paralel line to $AB$ from $D$. It intersects $r$ in $E$. Now $BE = \sqrt{3}AB$. Draw a circle of radius $BE$ from $D$. It intersects $s$ in $F$. Draw the line $AF$ and a paralel to $AF$ from $D$. It intersects $AB$ in $O$.
Now $AO:OB = \sqrt{3}:\sqrt{2}$.

A: Draw an isosceles rectangle triangle. Carry the hypothenuse on a side. If the hypothenuse is $\sqrt2$, the new segment is $\sqrt3$.

By Thales' theorem you can divide any segment in this ratio.
A: Picture first. Wordy description to follow.
We seek a point $C$ that divides $\overline{AB}$ into segments in proportion $\sqrt{2}:\sqrt{3}$.


Construct the line perpendicular to $\overleftrightarrow{AB}$ at $A$. (Note: The perpendicular at any point would work, but this gives me a named point to work with.)
Using any unit you like, mark-off eight congruent segments on the perpendicular. Let $C^{\prime\prime}$, $A^\prime$, and $B^\prime$ be the second, sixth, and eighth constructed endpoints of those segments, as shown in the diagram. Construct semicircles with diameter $\overline{AA^\prime}$ and $\overline{BB^\prime}$. (Note that the centers of these semicircles are conveniently among the constructed endpoints; that's actually why we constructed so many of them!)
Construct the perpendicular to $\overleftrightarrow{AA^\prime}$ at $C^{\prime\prime}$, and let $A^{\prime\prime}$ and $B^{\prime\prime}$ be the points where this line meets the semicircles. $\triangle AA^\prime A^{\prime\prime}$ is therefore a right triangle from which we derive the proportion
$$\frac{|\overline{AC^{\prime\prime}}|}{|\overline{A^{\prime\prime} C^{\prime\prime}}|} = \frac{|\overline{A^{\prime\prime} C^{\prime\prime}}|}{|\overline{A^\prime C^{\prime\prime}}|} \;\to\; |\overline{A^{\prime\prime} C^{\prime\prime}}|^2 = |\overline{A C^{\prime\prime}}||\overline{A^{\prime}C^{\prime\prime}}| = 2 |\overline{A C^{\prime\prime}}|^2  \;\;\to\;\; |\overline{A^{\prime\prime}C^{\prime\prime}}| = \sqrt{2}\,|\overline{A C^{\prime\prime}}|$$
Likewise, $|\overline{B^{\prime\prime}C^{\prime\prime}}| = \sqrt{3}\,|\overline{AC^{\prime\prime}}|$, so that $C^{\prime\prime}$ divides $|\overline{A^{\prime\prime}B^{\prime\prime}}|$ into lengths in proportion $\sqrt{2} : \sqrt{3}$.
Consequently, if $X$ is the intersection of $\overleftrightarrow{AA^{\prime\prime}}$ and $\overleftrightarrow{BB^{\prime\prime}}$, then line $\ell := \overleftrightarrow{XC^{\prime\prime}}$ meets $\overline{AB}$ at the desired division point $C$. (If it happens that $\overleftrightarrow{AA^{\prime\prime}}$ and $\overleftrightarrow{BB^{\prime\prime}}$ are parallel, then take $\ell$ to be the line, through $C^{\prime\prime}$, parallel to them.) $\square$
Notes:


*

*We get the same $C$ if we construct each semicircle on the "other side" of the perpendicular line.

*If we construct both semicircles on the "same side" of the perpendicular line, then we obtain a point $C$ external to $\overline{AB}$ yet with $|\overline{AC}|:|\overline{BC}| = \sqrt{2}:\sqrt{3}$. (We get the same external $C$, regardless of which side is the "same side".)

*This strategy can be used to subdivide a segment into any proportion of the form $\sqrt{m} : \sqrt{n}$.
A: The key is to remember that in any triangle, if you bisect an angle it cuts the opposite side into the same proportion as the sides of the angle.  (Source: http://hotmath.com/hotmath_help/topics/triangle-angle-bisector-theorem.html).  So to cut a segment into the ratio $\sqrt{2}:\sqrt{3}$, it is enough to bisect a triangle whose sides have that ratio.
Start with the given segment $AB$.  Using that as a base, construct an isosceles right triangle (I assume you know how to construct perpendiculars.)  Then the hypotenuse $AC$ of that triangle has length equal to $AB\sqrt{2}$.
Now construct right triangle $ABD$ as shown below.  It is easy to see that $BD=AB\sqrt{2}$ and $AD=AB\sqrt{3}$.  So if you bisect $\angle ADB$, it will cut $AB$ into the desired ratio. 

A: For ease of exposition, let's assume $|AB|=1$.  Now draw a perpendicular to $AB$ at $B$ and let $C$ and $D$ be points on the perpendicular with $C$ between $B$ and $D$ and $|BC|=|CD|=1$.  Next draw circles of radius $|BD|=2$ centered at $B$ and $D$, and let $E$ be a point of intersection of those two circles.  Note that $\triangle BDE$ is equilateral, with sides of length $2$, and $C$ is the midpoint of the side $BD$, so $|CE|=\sqrt3$.
Now draw a circle of radius $|CE|=\sqrt3$ centered at $C$, and then extend $AC$ until it intersects that circle at a point $F$, with $C$ between $A$ and $F$.  Note that $|AC|=\sqrt2$ and $|CF|=\sqrt3$.
Finally, construct the line through $C$ parallel to $BF$.  (Constructing parallel lines is a standard straightedge-and-compass procedure.)  This line intersects $AB$ at a point $P$, and we have 
$$|AP|:|PB|=|AC|:|CF|=\sqrt2:\sqrt3$$
Added later:  It occurs to me it might be worth counting the total number of distinct straightedge-and-compass steps involved.  Assuming we start with the extended line $AB$ (not just the line segment), constructing the perpendicular at $B$ can be done with $3$ circles and $1$ line; if you do it carefully (by first drawing the circle centered at $B$ of radius $|AB|$), you get the point $C$.  It then takes $1$ more circle to get $D$ and $2$ circles to get $E$.  To get $F$ takes a circle and a line.  Finally, constructing the parallel to $FB$ through $C$ can be done with $3$ circles and $2$ lines:  draw the line $FB$, draw the circle centered at $F$ of radius $|CF|$, intersecting $FB$ at $Q$, followed by circles centered at $C$ and $Q$ of radius $|CF|=|QF|$, which intersect at $R$, and finally the line $CR$, which is parallel to $FB$ (and intersects $AB$ at the desired point $P$).  Thus the entire construction involves a total of $10$ compass steps and $4$ straightedge steps.  (If all you have to start with are the two points $A$ and $B$, one more straightedge step, to construct the line $AB$, is needed.)
I'd be interested to know if any of the other answers' constructions get by with fewer steps.
A: To draw a line segment with ratio $a:b$ notice the total length of the line segment will be $n * (a + b),$ where $n$ is an arbitrary factor.  Set $$L = n*(a + b)$$ where $L$ is the length of the line segment to find $n.$ 
Does that give you enough to complete the solution?
A: Alternatively, let $AB$ be the line segment you want to divide with a point $P$ such that $\frac{|AP|}{|PB|}=\frac{\sqrt{2}}{\sqrt{3}}$.  Let $C$ be on the extension of $AB$ such that $B$ lies between $A$ and $C$ and that $|BC|=6\,|AB|$.  Draw the circle $\Gamma$ with diameter $AC$.  The line perpendicular to $AC$ at $B$ meets $\Gamma$ at $D$ and $E$.  Let $Q$ be the point on $DE$ such that $|DQ|=3\,|AB|$.  Then, the circle centered at $B$ with radius $BQ$ meets the interior of $AB$ at the required point $P$.
A: let AB be the given line Segment. Draw a line L1 from A which makes an angle 45 degrees, and line L2 from B which makes an angle 60 with AB. Let L1, L2 meet at C.
Now ABC is a triangle with angle A= 45, angle B = 60. Let the angular bisector of C (makes an angle 37.5 with either of AC,BC) meets AB at D. We know that $$AD/DB = AC/BC ..........(1)$$we also know that 
$$AC/sin(97.5) = CD/sin45$$$$ BC/sin(82.5) = CD/sin60$$ from the triangles CAD, CBD respectively
combining the above two equations

$$AC/BC = sin60/sin45 = (\sqrt{3}/2 )/(1/\sqrt{2}) = \sqrt{3}/\sqrt{2}$$ 
from (1)

$$AD/DB = \sqrt{3}/\sqrt{2}$$
Note that sin97.5 = sin 82.5
our line is divided by the point D in the required ratio.
