We all know about quadrilateral as a polygon with 4 sides. What is a Simson line of it? Is there a Simson line for a triangle? Yes !
Consider a 
and a point 
on the plane. Let the feet of perpendiculars from 
to 
be 
respectively. Then are collinear if and only if lies on the circumcircle of . If it happens that are collinear, then the line 
is called as the Simson line of point w.r.t . Try to prove this fact using the Inscribed Angle Theorem (i.e., An arc of a circle subtends equal angles at the other arc of the circle) and write your proof in the comments.
Let us now come back to quadrilaterals. Draw a quadrilateral 
so that opposite sides are not parallel and complete it by extending its sides to meet. Let 
.
Which circle should we take and which point should we take on it to draw the Simson line of the quadrilateral? Here is the catch, consider the circumcircles of 
, say 
. Let us prove that all these circles pass through a common point known as the Miquel Point of Quadrilateral.
Let 
be the intersection of 
and 
other than 
. Let us prove that also lies on 
and 
. By the Inscribed Angle Theorem on 
, 
. But we also know that 
and 
points 
are concyclic by the Inscribed Angle Theorem and hence lies on . Similarly, 
lies on as well. Hence, all the 
circles pass through this common point known as the Miquel point of Quadrilateral.
Now the Simson line is drawn w.r.t. the Miquel point by dropping perpendiculars from it to the sides 
of the quadrilateral. Of course, the feet of perpendiculars are collinear because the point lies on all four circumcircles and it will correspond to the Simson line of w.r.t all four triangles . Hence, this line will be called as Simson line of the quadrilateral ABCD.
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