triangle. Triangle D

the trilinear coordinates of the point D. All the elements of the pedal can be expressed in terms of these

trilinears and the sides of the triangle of reference ABC. Below is a compedium of such formulas.

Except the first and the last the other formulas give respective results for the other sides by cyclically

changing the coordinates.

[1] results by calculating the area of ABC in terms of (x,y,z).

[2,3] are shown in

CoordinatesBasics.html .

[4] is a consequence of the previous.

[5] is a calculation of the altitude of

a triangle.

[6, 7] were shown by Steiner as follows:

Because AD

(1/2)(AD

-DD

[7] is a consequence of [6] and [8] is proved in AreaOfPedal.html .

{E

[1] {E

[2] Points {D,E} are symmetric with respect to the circumcenter O of D

[3] Points {D,E} are isogonal conjugate with respect to ABC.

[4] Quadrangles DD

[5] Line AH, H being the intersection of lines {D

[6] Line D

[7] Quadrangles DD

[8] The perspectivity axis of point-perspective triangles {DD

center O of (c) and is orthogonal to AF.

Analogous statements hold for similar constructs with respect to the other vertices {B, C} of triangle ABC.

[1], [2] Consider the projections of O on the sides and identify E

[3] Angle chasing as indicated.

[4] Follows from [3] by showing the similarity of the corresponding triangles.

[5] H is on the radical axis of the circumcircles of the similar quadrangles (actually it is the radical

center of the three circles drawn). Since AD, AE are diameters of these circles, their radical axis is

orthogonal to line DE.

[6] Considering F to be the intersection point of {D

is the pole of AH with respect to the circle, thus the orthogonal OG to AJ (see [5]) passes through F.

[7] Follows from [6].

[8] The perspectivity axis is also diagonal of parallelogram DKEI, hence passes through the middle O

of DE. By an argument like that of [6] we deduce that H is the pole of line AF, hence OH is

orthogonal to AF.

The proof is an easy angle chasing as indicated by the picture. Pedal t

point P on the sides of t. Pedal t

the problem of inscribing a triangle into another triangle and the so-called pivots of this

inscription. Projecting namely the point P on the sides of ABC not orthogonally but at an angle (fi)

to the orthogonal we obtain triangles similar to the pedal. We obtain thus a family of similar

triangles A'B'C' of which the pedal one is the smallest in area (or perimeter).

This can be easily seen by observing that triangles {PA'B', PB'C', PC'A'} remain similar to

themselves and their sides opposite to P envelope a parabola whose one focus is P and the vertex

is the projection P' of P on a side of the pedal (see ParabolaNewton.html for details on this).

[2] Pedal of circumcenter, formed by the middles of the sides (medial triangle).

[3] Pedal of orthocenter of ABC: orthic triangle of ABC.

[4] Pedal of symmedian point of ABC: triangle similar to the one formed by the medians of ABC.

[2] Find properties of the pedal of the centroid.

[3] Characterize points P for which the lines joining their projections to opposite vertices are

concurrent (Rigby pedal cubic).

The above figure shows the Rigby pedal cubic of the triangle together with some points on it:

the incenter, circumcenter, orthocenter, Gergonne point, and de Longchamps point.

a point D. The

The

The trilinears of the traces {Q,R,S} can be computed easily from the trilinears (x,y,z) of P and the results

obtained in TrilinearsRelatedToOthers.html .

Analogously the trilinears of the projections {D,E,F} of P can be easily computed.

AreaOfPedal.html

ParabolaNewton.html

TrilinearsRelatedToOthers.html

Produced with EucliDraw© |