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.  

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.
