Can you construct an equilateral triangle? Assume each line has length 2.
Then bi-sect one of the angles (or lines)?
I.e. bi-sect the equilateral triangle?
Which of the lines can now be AD, because it has length root 3?
And then which end of it is best as A, in order to find B?
Actually... that requires a straight edge.
Ok then, with your compass mark a unit length from one end, A, then one further from the mark.
From the two marks, a point having distance of unit length from both.
How far is the new (third) mark from A? (Think, angles at a diameter.)
Mark a point, D, that same distance from A.
Yes, much simpler.
Compass point at A opened out to B, obtain a point C (on BA extended), so that CA=AB=1, CB=2 units.
Compass point at C opened out to B, draw a semi-circle.
Keep the same radius and draw a semi-circle centre B intersecting the last semi-circle (the one with centre C), at D'.
D' will be at a distance root 3 from A, so simply scribe the distance AD' onto the line AB to obtain the point D.
The proof of the result can be seen either from Pythagoras in the triangle CAD' or by making use of the intersecting chords theorem.