What is the Area of Triangle Lmn

In geometry, a
pedal triangle
is obtained by projecting a betoken onto the sides of a triangle.

More specifically, consider a triangle
ABC, and a indicate
P
that is not ane of the vertices
A, B, C. Drib perpendiculars from
P
to the three sides of the triangle (these may need to be produced, i.e., extended). Label
50,
M,
North
the intersections of the lines from
P
with the sides
BC,
AC,
AB. The pedal triangle is then
LMN.

If ABC is non an birdbrained triangle, P is the orthocenter then the angles of LMN are 180°−2A, 180°−2B and 180°−2C.^[1]

The location of the chosen point
P
relative to the chosen triangle
ABC
gives rise to some special cases:

• If
  P =
  orthocenter, and so
  LMN =
  orthic triangle.
• If
  P =
  incenter, then
  LMN =
  intouch triangle.
• If
  P =
  circumcenter, so
  LMN =
  medial triangle.

If
P
is on the circumcircle of the triangle,
LMN
collapses to a line. This is and so called the
pedal line, or sometimes the
Simson line
after Robert Simson.

The vertices of the pedal triangle of an interior point
P, as shown in the peak diagram, carve up the sides of the original triangle in such a mode as to satisfy Carnot’southward theorem:^[2]

$$


A

North

2


+
B

L

two


+
C

M

2


=
N

B

2


+
L

C

2


+
Grand

A

2


.


{\displaystyle AN^{2}+BL^{ii}+CM^{2}=NB^{2}+LC^{2}+MA^{ii}.}

Trilinear coordinates

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If
P
has trilinear coordinates
p :
q :
r, then the vertices
Fifty,G,N
of the pedal triangle of
P
are given by

• L = 0 : q + p
  cos C : r + p
  cos
  B
• 1000 = p + q
  cos
  C : 0 : r + q
  cos
  A
• N = p + r
  cos
  B : q + r
  cos
  A : 0

Antipedal triangle

[edit]

Ane vertex,
Fifty’, of the
antipedal triangle
of
P
is the betoken of intersection of the perpendicular to
BP
through
B
and the perpendicular to
CP
through
C. Its other vertices,
M
‘ and
N
‘, are constructed analogously. Trilinear coordinates are given by

• L’
  = − (q + p
  cos
  C)(r + p
  cos
  B) : (r + p
  cos
  B)(p + q
  cos
  C) : (q + p
  cos
  C)(p + r
  cos
  B)
• M’
  = (r + q
  cos
  A)(q + p
  cos
  C) : − (r + q
  cos
  A)(p + q
  cos
  C) : (p + q
  cos
  C)(q + r
  cos
  A)
• N’
  = (q + r
  cos
  A)(r + p
  cos
  B) : (p + r
  cos
  B)(r + q
  cos
  A) : − (p + r
  cos
  B)(q + r
  cos
  A)

For example, the excentral triangle is the antipedal triangle of the incenter.

Suppose that
P
does non lie on any of the extended sides
BC, CA, AB,
and let
P
^−one
announce the isogonal conjugate of
P. The pedal triangle of
P
is homothetic to the antipedal triangle of
P
⁻¹. The homothetic center (which is a triangle center if and but if
P
is a triangle centre) is the point given in trilinear coordinates past

ap(p + q
cos
C)(p + r
cos
B) : bq(q + r
cos
A)(q + p
cos
C) : cr(r + p
cos
B)(r + q
cos
A).

The product of the areas of the pedal triangle of
P
and the antipedal triangle of
P
^−one
equals the square of the area of triangle
ABC.

Pedal circle

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The pedal circle is divers as the circumcircle of the pedal triangle. Note that the pedal circle is not defined for points lying on the circumcircle of the triangle.

Pedal circle of isogonal conjugates

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For whatever bespeak

$$


P


{\displaystyle P}

not lying on the circumcircle of the triangle, it is known that

$$


P


{\displaystyle P}

and its isogonal cohabit

$$



P

⋆






{\displaystyle P^{\star }}

have a mutual pedal circle, whose middle is the midpoint of these ii points.^[iii]

References

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External links

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• Mathworld: Pedal Triangle
• Simson Line
• Pedal Triangle and Isogonal Conjugacy