Pedal Triangles
by Molly McKee
 
Construct any triangle ABC.
Let P be any point on the plane.
Construct the perpendiculars from point P to each side of ABC.
Find the points of intersection of these lines, R, S, and T.
The triangle formed by R, S, and T is the pedal triangle and P is the pedal point.
 
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Suppose that P, the pedal point, is the circumcenter of ABC.
 
Then P is the center of the circle which circumscribes ABC.
Therefore P is equidistant from all three vertices.
 
Suppose that P, the pedal point, is the incenter of ABC.
 
 
 
Then P is the center of the circle which is inscribed by ABC.
Therefore P is equidistant from all three vertices of the pedal triangle.
 
 
When P equals a vertice, say B, then the pedal triangle is a line.
 
Extend the line segment AB and find a line which passes through P and is perpendicular to AB.  Then the point of intersection between this line and AB will be B.  Similarly, extend the line segment BC and find a line which passes through P and is perpendicular to BC.  Then the point of intersection between this line and BC will be B.  Now extend line segment AC and find a line which passes through P and is perpendicular to AC.  We can see that the point of intersection between this line and AC will be S.  By definition, since B=P, the line BS is an altitude from vertex B of ABC.
 
Similarly, when A=P, the line AT is an altitude from vertex A of ABC.
When C=P, the line CR is an altitude from vertex C of ABC.
 
Suppose that P, the pedal point, is one of the vertices of ABC.
 
P as the Circumcenter
P as the Incenter
P as a Vertex
gsp Files about Pedal Triangleshttp://jwilson.coe.uga.edu/EMAT6680Su07/McKee/animation/circumcenter.htmlhttp://www.google.com/http://jwilson.coe.uga.edu/EMAT6680Su07/McKee/animation/incenter.htmlhttp://jwilson.coe.uga.edu/EMAT6680Su07/McKee/animation/vertex.html9.1_files/Assignment9.gsp9.1_files/xmovement.gspshapeimage_12_link_0shapeimage_12_link_1shapeimage_12_link_2shapeimage_12_link_3shapeimage_12_link_4