Example of 3 nonlinear points such that no square passes through all 3 points I know that for any 3 nonlinear point there exists circle which passes through that points.
I wanted to know is this fact true for square .
Can any one help me to give counterexample 
Any help will be appreciated
 A: It may help to consider the three points as vertices of a triangle.  
If the triangle is isosceles, with acute vertex angle at $C$, as in the first figure, draw a line through $B$ parallel to $AC$. Then draw $AE$, $CD$ perpendicular to that line, make $EF=GD=ED$ and complete square $EDGF$. This construction works also if $ABC$ is equilateral.
If the angle at $C$ is right, $AB$ is the diagonal of a square easy to complete.
If the angle at $C$ is obtuse, as in the second figure, draw a line through $C$ parallel to base $AB$, and complete the square as before.
Lastly, if triangle $ABC$ is scalene, as in the third figure, draw a line through the vertex of the greatest angle, parallel to the longest side of the triangle (here a line through $C$ parallel to $AB$), and complete the square. 
Thus it seems that, given three non-collinear points, a square can be constructed such that the three points lie on its perimeter.
A: Here is a proof avoiding to consider different figure cases.
Consider the figure below. For a given orientation $\theta$ and its orthogonal orientation $\theta+\pi/2$, we obtain 2 "minimal covering stripes" of triangle $ABC$, with red lines limits and blue lines limits resp. with widths $W_{\theta}$ and $W_{\theta+\pi/2}$ resp. 
If, for a given orientation $\theta$, $$S_{\theta}>S_{\theta+\pi/2}\tag{1}$$ (as is the case on the given figure), replacing $\theta$ by $\theta+\pi/2$ results in the reversal of order in (1). As functions $W_{\theta}$ and $W_{\theta+\pi/2}$ are continuous functions of angle $\theta$, there exists $\theta_0$ with $\theta<\theta_0<\theta+\pi/2$ where the widths are equal !

A: Consider the 3 random non co-linear points $A$, $B$ and $C$ :

Now if we join $AC$ and draw the $\perp$  from $B$ to$AC$

Now since $BD > AC$, we can construct a square of side length $BD$. So extend $AC$ such that $l(AE)=l(BD)$

Now if we draw square with side $AE$ such that it contains $B$ by drawing respective perpendiculars.

We have a square $AEFG$ (one of many) that passes through $A$, $B$, $C$ : 3 non-co-linear points.
So there will be at least 1 square that passes through 3 non-co-linear (or even co-linear points for that matter).
