The Simson line

Introduction. In the following JSP-applet construction you see a triangle ABC. Through a mobile point M, three green lines are drawn perpendicularly to the sides of the triangle. The intersections with the triangle's sides are denotes by P, Q and R, and joined by red lines.

1. Conjecture

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Problems

1. Find out where do lie those points M, for which the three points P, Q and R are collinear, i.e. are on the same line?
You may use tracing the point M when dragging. You may even force the system to not to draw too inaccurate points.

2. Based on your investigations, formulate a conjecture about the result.

2. Proof of the Simson theorem

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Notations. The line through A and B is denoted by AB. The unique oriented right angle of lines is denoted by d.
Rules. For any lines a, b, c, d, e, f and any points A, B, C and D:
 
(áa,bñ + ác,dñ ) + áe,fñ
= 
áa,bñ + (ác,dñ + áe,fñ) 
(1)
áa,añ
= 
0
(2)
áa,bñ = 0
Þ
a || b
(3)
a || b
Þ
áa,bñ = 0
(4)
a ^ b
Þ
áa,bñ = d
(5)
áa,bñ = d
Þ
a ^ b
(6)
áa,bñ
= 
- áb,añ
(7)
áa,bñ + ác,dñ
= 
ác,dñ + áa,bñ
(8)
áa,bñ + áb,cñ
= 
áa,cñ
(9)
A, B, C collinear
Û
AB = AC
(10)
A, B, C, D cocyclical
Û
áCA,CBñ = áDA,DBñ
(11)
áCA,CBñ = áDA,DBñ
Û
A, B, C, D cocyclical
(12)

Problem

Complete the proof by indicating the number of the rule or relation used at each step of the argument:
áPM,PCñ = d and áQM,QCñ = d because of   5 
P, Q, M and C are then cocyclical  because of  __ 
and so áMC,MPñ = áQC,QPñ (13)  because of  __

In the same way:

áPM,PBñ = d and áRM,RBñ = d because of  __ 
P, Q, M and B are then cocyclical  because of  __ 
and so áMP,MBñ = áRP,RBñ (14)  because of  __ 

We have:

áMC,MBñ = áMC,MPñ + áMP,MBñ because of  __ 
= áQC,QPñ + áRP,RBñ because of (13) and  __ 
= áAC,QPñ + áRP,ABñ because of  __ used twice
= (áAC,ABñ + áAB,QPñ) + áRP,ABñ because of  __ 
= áAC,ABñ + (áAB,QPñ + áRP,ABñ) because of  __
= áAC,ABñ + áRP,QPñ (15) because of  __ 

If M belongs to the circumcircle of ABC, then A, B, C and M are cocyclical and because of __ , we have:

áMC,MBñ = áAC,ABñ.
From (15) and from the rules 1, 7 and 8 we deduce:
áRP,QPñ = 0.
The lines RP and QP are then parallel because of __ . These two parallel lines go both through P, so they are equal and the points P, Q and R are collinear. This means that M belongs to the locus. So we have shown:

(circumcircle of ABC) Ì (locus of M).             (16)

If M belongs to the locus, P, Q and R are collinear and:

áRP,QPñ = 0     because of 
and so, since áa,bñ + 0 = áa,bñ:
áMC,MBñ = áAC,ABñ.
A, B, C and M are then cocyclical because of __ , and then M belongs to the circle going through A, B and C, so:

(locus of M) Ì (circumcircle of ABC).             (17)

Finally as a consequence of __  and __  we get:

(locus of M) = (circumcircle of ABC).             

q.e.d.