Global Journal of Advanced Research
on Classical and Modern Geometries
ISSN: 2284-5569, Vol.4, (2015), Issue 1, pp.44-54
PROPERTIES INVOLVING PEDAL TRIANGLES
SÁNDOR NAGYDOBAI KISS
A BSTRACT. We deduce the area of a pedal triangle in various forms in barycentric coordinates and we establish a relation between the area of the pedal triangles of isogonal
conjugates.
2010 Mathematical Subject Classification: 97G40, 97G70
Keywords and phrases: Pedal triangle, isogonal conjugates.
1. I NTRODUCTION
In this paper we deduce the area of a pedal triangle in various forms. We done the
condition that a triangle inscribed in the triangle of reference ABC to be a pedal triangle.
We will demonstrate that the symmetric points of the vertex of a pedal triangle with
respect to the midpoints of the sides of ABC triangle form a pedal triangle. Finally we
establish a relation between the area of the pedal triangles of isogonal conjugates.
We use the barycentric coordinates ([2], [3]) with respect to the ABC triangle: A =
(1, 0, 0), B = (0, 1, 0), C = (0, 0, 1). Let a, b, c be the length of sidelines BC, CA, AB
and R the circumradius of the ABC triangle. In the plan of the triangle ABC we consider a varying point M (Figure 1), with barycentric coordinates M = (u : v : w) so that
uvw 6= 0 and note with the symbol µ M the sum of coordinates of M : µ M = u + v + w.
abc
the twice of area
Let s be the semiperimeter (2s = a + b + c), σ the area, S = 2σ =
4R
1
1
of ABC triangle, S A = bc cos A = (− a2 + b2 + c2 ), SB = ca cos B = ( a2 − b2 + c2 ),
2
2
1 2
2
2
SC = ab cos C = ( a + b − c ) so that S A SB SC 6= 0. More we introduce the following
2
notation: Sθ · Sϑ = Sθϑ .
Definition 1. Let Ma , Mb , Mc be the orthogonal projections of the M point on the sidelines BC, CA, AB. The Ma Mb Mc triangle is called the pedal triangle of M [1] (Figure 1).
I would like to dedicate this article to Professor Béla Orbán, on the occasion of his 85th birthday.
44
Properties Involving Pedal Triangles
A
Mc
.
..
Mb
M
.
Ma
B
C
Figure 1
Note with the symbol σ[ A1 A2 . . . An ] the area of a polygon A1 A2 . . . An .
In this paper we use the following conditioned trigonometric identities [2]:
b2 c2 − S2A = c2 a2 − S2B = a2 b2 − SC2 = S2 ,
(1.1)
a2 S A + SBC = b2 SB + SCA = c2 SC + S AB = S2 ,
(1.2)
sin2 A + sin2 B + sin2 C = 2(1 + cos A cos B cos C ),
(1.3)
SBC + SCA + S AB = S2 ,
(1.4)
a2 S A + b2 SB + c2 SC = 2S2 ,
(1.5)
2 ( a 2 b 2 c 2 + S A S B SC ) = ( a 2 + b 2 + c 2 ) S 2 ,
(1.6)
2 ( S A + S B + SC ) = a 2 + b 2 + c 2 .
(1.7)
The demonstration of these relations are the following:
b2 c2 − S2A = b2 c2 − b2 c2 cos2 A = b2 c2 sin2 A =
a2 b2 c2
= S2 ,
4R2
a2 S A + SBC = a2 bc(cos A + cos B cos C ) = a2 bc · sin B sin C =
a2 b2 c2
= S2 ,
4R2
1 − cos 2A 1 − cos 2B
+
+ 1 − cos2 C
2
2
2 − 2 cos( A − B) cos( A + B) − cos2 C = 2 + 2 cos C [cos( A − B) − cos C ]
π − 2B
π − 2A
sin
= 2 + 2 cos A cos B cos C,
= 2 + 2 cos C sin
2
2
sin2 A + sin2 B + sin2 C =
SBC + SCA + S AB = abc( a cos B cos C + b cos C cos A + c cos A cos B)
= abc(c cos C + c cos A cos B) = abc2 sin A sin B =
a2 b2 c2
= S2 ,
4R2
2( a2 b2 c2 + S A SB SC ) = 2a2 b2 c2 (1 + cos A cos B cos C ) = ( a2 + b2 + c2 )S2 ,
2 ( S A + S B + SC ) = − a 2 + b 2 + c 2 + a 2 − b 2 + c 2 + a 2 + b 2 − c 2 = a 2 + b 2 + c 2 .
45
Sándor Nagydobai Kiss
2. VARIOUS
FORMS OF THE AREA OF PEDAL TRIANGLES IN BARYCENTRIC
COORDINATES
Theorem 1. Let ABC be a triangle and M = (u : v : w) a point with barycentric coordinates with respect to the triangle ABC. The area of the pedal triangle Ma Mb Mc in barycentric
coordinates is
4σ3
| a2 vw + b2 wu + c2 uv|
a2 b2 c2 µ2M
σ
| a2 vw + b2 wu + c2 uv|.
=
4R2 µ2M
σ [ M a Mb Mc ] =
(2.1)
Proof. The equation of the line, which passes
through the point ( x ′ : y′ : z′ ) and is per
x y z
pendicular to the line lx + my + nz = 0 is x ′ y′ z′ = 0, where
µ a µb µc µ a = la2 − mSC − nSB , µb = mb2 − nS A − lSC , µc = nc2 − lSB − mS A
(see [2]).
The equations of the line MMa , MMb and MMc are
x
y
z
2
2
u
2 v w = 0 ⇔ (vSB − wSC ) x − (wa + uSB )y + (va + uSC )z = 0,
− a SC S B x
y
z u
v
w = 0 ⇔ (wb2 + vS A ) x + (wSC − uS A )y − (ub2 + vSC )z = 0,
SC − b 2 S A x
y
z
u
v
w = 0 ⇔ −(vc2 + wS A ) x + (uc2 + wSB )y + (uS A − vSB )z = 0.
S B S A − c2 The barycentric coordinates of the Ma , Mb , Mc points are
Ma = (0 : va2 + uSC : wa2 + uSB ),
Mb = (ub2 + vSC : 0 : wb2 + vS A ),
Mc = (uc2 + wSB : vc2 + wS A : 0).
Consequently, the absolute barycentric coordinates of the Ma , Mb , Mc points are
uSC w
uSB
v
va2 + uSC wa2 + uSB
,
+ 2
,
+ 2
= 0,
,
Ma = 0,
a2 µ M
a2 µ M
µM
a µM µM
a µM
2
ub + vSC
wb2 + vS A
vSC
w
vS A
u
Mb =
=
,
, 0,
+ 2
, 0,
+ 2
b2 µ M
b2 µ M
µM
b µM
µM
b µM
2
wSB
v
wS A
u
uc + wSB vc2 + wS A
,
,0 =
+ 2
,
+ 2
,0 .
Mc =
c2 µ M
c2 µ M
µM
c µM µM
c µM
46
Properties Involving Pedal Triangles
In absolute barycentric coordinates the area of the triangle determined by the points
M1 =
(u1 , v1 , w1), M2 = (u2 , v2 , w2 ), M3 = (u3 , v3 , w3 ) is σ[ M1 M2 M3 ] = |∆|σ, where
u 1 v 1 w1 ∆ = u2 v2 w2 (see [2]). The area of Ma Mb Mc triangle is σ[ Ma Mb Mc ] = |∆|σ, where
u 3 v 3 w3 va2 + uSC wa2 + uSB 2 0
1
∆ = 2 2 2 3 ub + vSC
0
wb2 + vS A a b c µ M uc2 + wS vc2 + wS
0
B
A
=
=
1
a2 b2 c2 µ2M
1
a2 b2 c2 µ2M
2
a va2 + uSC wa2 + uSB 2
2 + vS b
0
wb
A
c2 vc2 + wS A
0
[− a2 (wb2 + vS A )(vc2 + wS A )+ b2 (vc2 + wS A )(wa2 + uSB )
+ c2 (va2 + uSC )(wb2 + vS A )]
=
=
1
a2 b2 c2 µ2M
[ a2 vw(b2 c2 − S2A )+ b2 wu(c2 SC + S AB )+ c2 uv(b2 SB + SCA )]
S2
4σ2
2
2
2
(
a
vw
+
b
wu
+
c
uv
)
=
( a2 vw + b2 wu + c2 uv).
a2 b2 c2 µ2M
a2 b2 c2 µ2M
So
σ [ M a Mb Mc ] =
=
4σ3
| a2 vw + b2 wu + c2 uv|
a2 b2 c2 µ2M
σ
4R2 µ2M
| a2 vw + b2 wu + c2 uv|.
Remark 1. An other form of the area of the pedal triangle is the following:
σ [ M a Mb Mc ] =
=
σ
| a2 vw + b2 wu + c2 uv|
4R2 µ2M
σ
|vw · sin2 A + wu · sin2 B + uv · sin2 C |
µ2M
(2.2)
u
u
v
v
w
w
=
, yM =
=
, zM =
=
µM
u+v+w
µM
u+v+w
µM
u+v+w
be the absolute barycentric coordinates of the M point. In these coordinates the area of
the pedal triangle Ma Mb Mc is
Remark 2. Let x M =
4σ3
| a2 y M z M + b2 z M x M + c2 x M y M |
a2 b2 c2
σ
| a2 y M z M + b2 z M x M + c2 x M y M |
=
4R2
σ [ M a Mb Mc ] =
= σ|y M z M · sin2 A + z M z M · sin2 B + x M y M · sin2 C |.
47
(2.3)
Sándor Nagydobai Kiss
If Xi is a triangle center [3], note the orthogonal projections of Xi on the sidelines BC, CA, AB
with Xai , Xbi , Xci , respectively. So the pedal triangle of Xi is Xai Xbi Xci . Now we calculate
the area of pedal triangle of Xi for i ∈ {1, 2, 3, 4}.
2.1. The point X1 = ( a : b : c) is the incenter of the ABC triangle. Its pedal triangle is
called the intouch triangle:
σ[ Xa1 Xb1 Xc1 ] =
4σ3
2σ3 (s − a)(s − b)(s − c)
2
2
2
|
a
bc
+
b
ca
+
c
ab
|
=
=
.
a2 b2 c2 4s2
abcs
2R
2.2. The point X2 = (1 : 1 : 1) is the centroid of ABC triangle. The area of its pedal
triangle is
σ
σ 1 2
· | a + b2 + c2 | = | sin2 A + sin2 B + sin2 C |
4R2 9
9
S
= (1 + cos A cos B cos C ).
9
σ[ Xa2 Xb2 Xc2 ] =
2.3. The point X3 = ( a2 S A : b2 SB : c2 SC ) is the circumcenter of ABC triangle. Its pedal
triangle is called the medial triangle:
σ[ Xa3 Xb3 Xc3 ] =
a2 b2 c2
σ
σ3
4σ3
·
= .
|
S
+
S
+
S
|
=
BC
CA
AB
2
2
2
2
4
a b c
S
4
4S
2.4. The point X4 = (SBC : SCA : S AB ) is the orthocenter of ABC triangle. Its pedal triangle
is called the orthic triangle:
4σ3
|S S S |
· A B4 C | a2 S A + b2 SB + c2 SC |
2
2
2
a b c
S
8σ3
a2 b2 c2 | cos A cos B cos C |
= 2 2 2·
a b c
S2
= S| cos A cos B cos C |.
σ[ Xa4 Xb4 Xc4 ] =
Remark 3. If ABC is an acute-angled triangle, then comparing the areas of pedal triangles Xa2 Xb2 Xc2 and Xa4 Xb4 Xc4 , we obtain
2σ[ ABC ] = 9σ[ Xa2 Xb2 Xc2 ] − σ[ Xa4 Xb4 Xc4 ].
(2.4)
3. C ONDITIONS NEEDED IN ORDER THAT AN INSCRIBED TRIANGLE IN A TRIANGLE TO
BE A PEDAL TRIANGLE
A Ta Tb Tc triangle is inscribed in an ABC triangle if Ta lies on BC, Tb lies on CA and Tc
lies on AB. Consider the Ta , Tb , Tc points with absolute barycentric coordinates: Ta =
(0, α, 1 − α), Tb = (1 − β, 0, β), Tc = (γ, 1 − γ, 0) (Figure 2). What is the condition needed
so that Ta Tb Tc triangle to be a pedal triangle?
48
Properties Involving Pedal Triangles
A
tc
.
Tc
tb
.
Tb
B
.
Ta
C
ta
Figure 2
Theorem 2. The Ta Tb Tc triangle inscribed in a ABC triangle is a pedal triangle if and only if
a2 α + b2 β + c2 γ =
1 2
( a + b 2 + c 2 ) = S A + S B + SC .
2
(3.1)
Proof. Let t a be the perpendicular to BC through Ta , tb the perpendicular to CA through
Tb , tc the perpendicular to AB through Tc (Figure 2). The equations of these lines are
x
y
z
α 1 − α = 0 ⇔ ( a2 α − SC ) x − a2 (1 − α)y + a2 αz = 0,
ta : 0
− a 2 SC
SB x
y
z 0
β = 0 ⇔ b2 βx + (b2 β − S A )y − b2 (1 − β)z = 0.
tb : 1 − β
2
SC
−b S A x
y
z 0 = 0 ⇔ −c2 (1 − γ) x + c2 γy + (c2 γ − SB )z = 0.
tc : γ 1 − γ
SB
SA
− c2 The t a , tb , tc lines are concurrent if and only if
2
a α − SC
− a2 (1 − α )
a2 α
2
2
2
b β
b β − S A −b (1 − β) = 0 ⇔
− c2 (1 − γ )
c2 γ
c2 γ − S B S B − a2 (1 − α ) − a2 SA
b2 β − S A
SC = 0 ⇔
− c2
c2 γ
SB 49
Sándor Nagydobai Kiss
⇔ (b2 β − S A )S2B − c2 a2 [γS A − (1 − α)SC + (b2 β − S A )]
−c2 γSB SC + a2 (1 − α)S A SB = 0 ⇔
(b2 β − S A )S2B − c2 a2 [γS A + αSC − b2 (1 − β)]
−c2 γSB SC + a2 (1 − α)S A SB = 0 ⇔
a2 (c2 SC + S A SB )α + b2 (c2 a2 − S2B ) β + c2 ( a2 S A + SB SC )γ
= a 2 b 2 c 2 + S A S B SC ⇔
a 2 S 2 α + b 2 S 2 β + c 2 S 2 γ = a 2 b 2 c 2 + S A S B SC ⇔
1
a2 α + b2 β + c2 γ = ( a2 + b2 + c2 ) ⇔
2
2
2
2
a α + b β + c γ = S A + S B + SC .
Remark 4. The conditions (3.1) can be written in the following equivalent forms:
a 2 α + b 2 β + c 2 γ = a 2 + S A = b 2 + S B = c 2 + SC ,
or
a2 α + b2 β + c2 γ = a2 + bc cos A = b2 + ca cos B = c2 + ab cos C,
or
a2 (2α − 1) + b2 (2β − 1) + c2 (2γ − 1) = 0.
Introduce the following notation:
E(α, β, γ) = a2 (2α − 1) + b2 (2β − 1) + c2 (2γ − 1).
Now we suppose that the t a , tb , tc lines are concurrent and let T = t a ∩ tb ∩ tc . We determine the coordinates of the T point, solving the following system with Cramer rule:
(
( a2 α − SC ) x − a2 (1 − α)y = − a2 αz
−c2 (1 − γ) x + c2 γy = −(c2 γ − SB )z
2 (1 − α ) a2 α − S
−
a
C
δ= 2
= c2 ( a2 α + S B γ − a2 ).
− c (1 − γ )
c2 γ
− a2 αz
− a2 (1 − α)
2
2
δx = 2
= − a (SB α + c γ − SB )z,
−(c γ − SB )z
c2 γ
2
a α − SC
− a2 αz = −( a2 S A α − c2 SC γ + SB SC )z.
δy = 2
−c (1 − γ) −(c2 γ − SB )z
So the barycentric coordinates of the T point are
T = [ a2 (SB α + c2 γ − SB ) : a2 S A α − c2 SC γ + SB SC : −c2 ( a2 α + SB γ − a2 )].
Note briefly the coordinates of T with x T , y T , z T . Their sum is:
x T + y T + z T = a 2 ( S B α + c 2 γ − S B ) + a 2 S A α − c 2 SC γ + S B SC
− c2 ( a2 α + SB γ − a2 ) = c2 a2 − S2B = S2 .
50
Properties Involving Pedal Triangles
Now we give these coordinates in another form:
a2 α + b2 β + c2 γ
2
2
2
2
xT = a (SB α + c γ − SB ) = a SB α + c γ −
SB
S A + S B + SC
a2
[(SB α + c2 γ)( a2 + S A ) − ( a2 α + b2 β + c2 γ)SB ]
S A + S B + SC
a2
=
( S A S B α − b2 S B β + b2 c2 γ ),
S A + S B + SC
=
y T = a 2 S A α − c 2 SC γ + S B SC = a 2 S A α − c 2 SC γ +
a2 α + b2 β + c2 γ
S B SC
S A + S B + SC
1
[( a2 S A α − c2 SC γ)(b2 + SB )+ ( a2 α + b2 β + c2 γ)SB SC ]
S A + S B + SC
b2
( c 2 a 2 α + S B SC β − c 2 SC γ ) ,
=
S A + S B + SC
a2 α + b2 β + c2 γ 2
2
2 2
2
2
zT = −c ( a α + SB γ − a ) = −c a α + SB γ −
a
S A + S B + SC
=
c2
[−( a2 α + SB γ)(c2 + SC ) + ( a2 α + b2 β + c2 γ) a2 ]
S A + S B + SC
c2
(− a2 S A α + a2 b2 β + SC S A γ).
=
S A + S B + SC
=
Consequently, the absolute barycentric coordinates of the T point are
T=
1
( S A + S B + SC ) S 2
[ a2 ( S A S B α − b2 S B β + b2 c2 γ ),
b2 (c2 a2 α + SB SC β − c2 SC γ), c2 (− a2 S A α + a2 b2 β + SC S A γ)]
Let A′ , B′ , C ′ be the midpoints of the side BC, CA, AB and the pairs of points ( Ta , Ta′ ),
( Tb , Tb′ ), ( Tc , Tc′ ) to be symmetric with respect to the A′ , B′ , C′ midpoints (Figure 3).
A
. .
Tb
Tc
.
.
B'
C'
T'c
.
O
.
T'b
B
Ta
A'
Figure 3
51
.
.
C
T'
Sándor Nagydobai Kiss
The absolute barycentric coordinates of the Ta′ , Tb′ , Tc′ points are Ta′ = (0, 1 − α, α), Tb′ =
( β, 0, 1 − β), Tc′ = (1 − γ, γ, 0). Whether the triangle Ta′ Tb′ Tc′ is a pedal triangle as the
triangle Ta Tb Tc ?
Theorem 3. The symmetries of the vertex in a pedal triangle with respect to the midpoints of
sides of ABC reference triangle are the vertex of a pedal triangle.
Proof. The triangle Ta Tb Tc is a pedal triangle, so
E(α, β, γ) = a2 (2α − 1) + b2 (2β − 1) + c2 (2γ − 1) = 0.
The triangle Ta′ Tb′ Tc′ is pedal triangle, if and only if E(1 − α, 1 − β, 1 − γ) = 0, which is
true, since E(1 − α, 1 − β, 1 − γ) = − E(α, β, γ) = 0.
Definition 2. If the pairs of ( Ta , Ta′ ), Tb , Tb′ ), ( Tc , Tc′ ) points are symmetric with respect to
the A′ , B′ , C ′ midpoints of the sides BC, CA, AB then the two Ta Tb Tc and Ta′ Tb′ Tc′ triangles
we call symmetric pedal triangles.
Let Ta′ Tb′ Tc′ be the pedal triangle of the point T ′ . The absolute barycentric coordinates of
T ′ are:
2
1
T′ =
a [S A SB (1 − α) − b2 SB (1 − β) + b2 c2 (1 − γ)],
2
( S A + S B + SC ) S
b2 [c2 a2 (1 − α) + SB SC (1 − β) − c2 SC (1 − γ)],
c2 [− a2 S A (1 − α) + a2 b2 (1 − β) + SC S A (1 − γ)] .
Theorem 4. The area of symmetric pedal triangles Ta Tb Tc and Ta′ Tb′ Tc′ are equal and the T and
T ′ points are symmetric with respect to O, the circumcenter of the ABC triangle.
Proof. We calculate the area of Ta Tb Tc and Ta′ Tb′ Tc′ triangles:
σ[ Ta Tb Tc ] = |∆1 |σ, where
0
α
1 − α
0
β = αβγ + (1 − α)(1 − β)(1 − γ),
∆1 = 1 − β
γ
1−γ
0 σ[ Ta′ Tb′ Tc′ ] = |∆2 |σ, where
0
1−α
α 0
1 − β = αβγ + (1 − α)(1 − β)(1 − γ).
∆2 = β
1 − γ
γ
0 We will demonstrate that x T + x T ′ = 2x0 , where
a 2 S A b 2 S B c 2 SC
O = ( x0 , y0 , z0 ) =
,
,
:
2S2 2S2 2S2
x T + x T ′ = 2x0 ⇔ x T + x T ′ =
a2 S A
⇔
S2
⇔ a 2 ( S A S B − b 2 S B + b 2 c 2 ) = a 2 S A ( S A + S B + SC ) ⇔
⇔ − b 2 S B + b 2 c 2 = S A ( S A + SC ) ⇔
⇔ − b2 S B + b2 c2 = S A b2 ⇔ S A + S B = c2 ,
52
Properties Involving Pedal Triangles
which is true (Figure 3). Similarly it is possible to demonstrate that y T + y T ′ = 2y0 and
z T + z T ′ = 2z0 .
4. R ELATION
BETWEEN THE AREA OF PEDAL TRIANGLES OF ISOGONAL CONJUGATES
The isogonal conjugate N of an M point in the plane of the ABC triangle is constructed
by reflecting the lines AM, BM and CM of the angle bisectors at A, B and C. The three
reflected lines concur at the isogonal conjugate [4], [5]. Consequently, the barycentric
coordinates
2 of2 N are
b
c2
a
:
:
= ( a2 vw : b2 wu : c2 uv). Note with the symbol µ N the sum of the
N =
u
v
w
a2 vw
b2 wu
c2 uv
coordinates of N : µ N = a2 vw + b2 wu + c2 uv. Let x N =
, yN =
, zN =
µN
µN
µN
be the absolute barycentric coordinates of N (Figure 4).
A
.
Mb
Mc
..
M
.
Nb
Nc
.
.
.
.
Ma
Na
N
B
C
Figure 4
Theorem 5. The following relation stands:
| x N y N z N | · σ[ Ma Mb Mc ] = | x M y M z M | · σ[ Na Nb Nc ].
Proof. The area of the pedal triangle Ma Mb Mc is:
σ [ M a Mb Mc ] =
4σ3 |µ N |
4σ3
2
2
2
.
|
a
vw
+
b
wu
+
c
uv
|
=
a2 b2 c2 |µ2M |
a2 b2 c2 µ2M
53
(4.1)
Sándor Nagydobai Kiss
The area of the Na Nb Nc pedal triangle is:
σ[ Na Nb Nc ] =
=
4σ3
| a2 b2 wu · c2 uv + b2 c2 uv · a2 vw + c2 a2 vw · b2 wu|
2
2
2
2
a b c µN
4σ3
· a2 b2 c2 |uvw| |µ M |
a2 b2 c2 µ2N
4σ3 |µ N | |µ3M | 2 2 2
·
a b c |uvw|
a2 b2 c2 µ2M |µ3N |
2
a vw b2 wu c2 uv |µ3M |
·
·
·
· σ [ M a Mb Mc ] .
= µN
µN
µ N |uvw|
=
From this, the relation (4.1) follows.
With the formula (4.1) we can determine the area of Na Nb Nc pedal triangle, if we know
the area of pedal triangle Ma Mb Mc and vice versa.
For example: we calculate the area of pedal triangle of the symmedian point
a2
b2
c2
,
K=
,
,
a2 + b2 + c2 a2 + b2 + c2 a2 + b2 + c2
1 1 1
.
, ,
which is the isogonal conjugate of the centroid G =
3 3 3
So
| xG yG zG | · σ[Ka Kb Kc ] = | xK yK zK | · σ[ Ga Gb Gc ] ⇔
⇔
a2 b2 c2
1
· σ [ K a Kb Kc ] = 2
· σ[ Ga Gb Gc ] ⇔
27
( a + b2 + c2 )3
⇔ σ [ K a Kb Kc ] =
⇔ σ [ K a Kb Kc ] =
27a2 b2 c2
4σ3
1
·
· ( a2 + b2 + c2 ) ⇔
( a2 + b2 + c2 )3 a2 b2 c2 9
12σ3
.
( a2 + b2 + c2 )2
R EFERENCES
[1] Coxeter, H. S. M. and Greitzer, S. L., Pedal Triangles § 1.9 in Geometry Revisited. Washington DC: Math.
Assoc. Amer., 1967, 22–26.
[2] Kiss, Sándor, Comparative analysis of coordinate geometry methods, Ed. Did. Ped. Bucureşti, 2008 (in Hungarian).
[3] Kimberling, Clark, Encyclopedia of Triangle Centers – ETC, http://faculty.evansville.edu/ck6/encyclopedia/ETC.html
[4] Yiu, Paul, Introduction to the Geometry of the Triangle, http://math.fau.edu/yiu/GeometryNotes020402.pdf
[5] Sigur, S., Where are the Conjugates? Forum Geom. 5 (2005), 1–15,
http://forumgeom.fau.edu/FG2005volume5/FG200501index.html
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