How to draw a square through 4 points?

Question

Draw a square ABCD. Select one point on each of the lines AB, BC, CD, DA (not at the vertices). Remove the square, so that you can see only the 4 points. Is it possible to reconstruct the square using a ruler and a compass?

Answer 1

I think I can suggest a simpler solution than NorthStar's. I'll assume the given points P1, P2, P3 and P4 are ORDERED: let A, B, C, D are consequent vertexes of the required square, P1 is on AB, P2 - on BC, P3 - on CD and P4 - on DA. Please follow the link to see how it is looking like:
http://i219.photobucket.com/albums/cc101/Andrey_Dyukmedzhiev/Square-1.gif

Part 1 (Analysis): let P1' is an orthogonal projection of P1 on CD, P2' - projection of P2 on DA and let's draw a line through P2, perpendicular to P1P3, intersecting DA in a point, say Q. Then triangles P1P1'P3 and P2P2'Q are congruent. Hence |P2Q| = |P1P3| and we have 2 points on DA, that will help in the construction below.

Part 2 (Construction): Connect P1 with P3, draw a line, perpendicular to P1P3 through P2, take on it a point Q such that |P2Q| = |P1P3| (see details below). The line P4Q is one of the required sides. The opposite is parallel to P4Q through P2, the rest is easy.

Part 3 (Proof): It's almost obvious that the constructed square is the required one.

Part 4 (Details): There are 2 ways to construct the point Q above (in each of both possible direction from P2), so in general, the problem will have 2 solutions, but, if Q ≡ P4 above, the solutions will be infinitely many (e.g. if P1P2P3P4 is a square - as shown on the picture).
The above assumes P1 and P3 to be on opposite sides of the square, otherwise we obtain 2*3 = 6 solutions /or, eventually infinitely many/.

Answer 2

Yes, it is possible. In fact, the question can be more general than that. The points can be one on each side or an extension of that side.

Here is a link to the Math Forum - Ask Dr. Math, which answers the question and provides a diagram.

http://mathforum.org/library/drmath/view/54675.html

Answer 3

yeh, pick the mid points of each side, u can do

connect 2 opposite points to make a cross and measure from what point the lines cross and end. then at the end of the line draw half of a whole line (distance u just measured) at 90 degrees from the adjacent line, repeat the same thing for all the line segments.