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How can I construct a square using only a pencil and a compass, i.e. no ruler.

Given: a sheet of paper with $2$ points marked on it, a pencil and a compass.

Aim: plot $2$ vertices such that the $4$ of them form a square using only a compass.

P.S.: no cheap tricks involved.

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    I think OP means constructing four points that can be the vertices of a square, using only a compass. Interesting question... – 2017-02-05
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    http://www.cut-the-knot.org/do_you_know/compass9.shtml – 2017-02-05
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    @Aretino this link is an excellent answer! Thanks – 2017-02-05
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    This question is better worded as constructing a square with *compass and straightedge*. The difference between a ruler and a straightedge is that the former is graduated by distance while the latter is not. However, measuring distances doesn't actually help you construct a square, so the problem might as well allow use of a ruler. – 2017-02-05
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    @DepressedDaniel: I agree that it's better to use the term "straightedge", but what the OP wants is to construct a square *without* using a straightedge. It's more difficult than a classical "compass and straightedge" construction. – 2017-02-05
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    Note that [the Mohr–Mascheroni theorem](https://en.wikipedia.org/wiki/Mohr%E2%80%93Mascheroni_theorem) implies that this is possible. – 2017-02-05

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The only difficulty is to create Perpendicular lines. One possible way could be, if you are precise enough:

lets suppose the vertices are (P1,P2,P3,P4), where P1 and P2 are given two points.

Join the two points(P1,P2) and extend on both sides.

To draw a perpendicular at point(P1) on the line: Mark two points on the line equidistant from P1 on both sides, then place your compass at the marked points, there after create arcs using your compass such that they intersect.

joining the intersection of arcs and P1 would give you a perpendicular line on the given line.

Since you have created a perpendicular line, then it's not difficult to create a Square from it.

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    It says no ruler! How are you going to join the two points $p1,p2$, let alone extending them? – 2017-02-05
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    @Seyed I suppose he can draw a perfect straight line using his hand. – 2017-02-05
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    Very well !!!!! – 2017-02-05
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On way to do this is to start by constructing the middle point of your segment like this:

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Then you will easily have the middle of your square like this:

enter image description here

And you know can have a circonscrit circle for your square, and constructing the square is finally possible!

Edit.

Since you asked for it, I have made a few more drawings to illustrate how to construct the point $O$ from where we left it.

enter image description here enter image description here enter image description here enter image description here enter image description here

Where the last circle has for radius $CF$ and for center $A$.

Edit 2.

Since more details were requested, here is how to finish the proof once $O$ has been constructed.

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    How do you construct point "O" in the last figure? – 2017-02-05
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    @Aretino I edited to specify that particular point. – 2017-02-05
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    But I still don't understand how would you join the points ? – 2017-02-05
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    @A---B You don't need to. When you have $O$, you can draw a circle that will pass trough every angle of your square, and it is now easy to construct the square. – 2017-02-05
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    I understand your construction but we still need to join the vertices of square on the circumference of the circle ? – 2017-02-05
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    @A---B I did more drawings, and I hope it answers what you are asking me. – 2017-02-05
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    your answer is certainly correct,but if you had noted that that we can consrtuct a square of side BC trivially with vertex O.So what if we do that right from the beginning.i.e given a side AB,we can construct a square with side AB/2.so lets just double AB – 2017-02-07
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    @TANUJ I think you are right, my answer is certainly not the fastest one! – 2017-02-07
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The key to solve this problem is how to construct $\sqrt{2}$. enter image description here