Theorem 4.22
(Brocard)
Let ABCD be a quadrilateral inscribed in a circle with
center
O
and let
E
=
AB
∩
CD
,
F
=
AD
∩
BC
,
and
J
=
AC
∩
BD
.
Then
O
is the orthocenter of the triangle EFJ
.
Theorem 4.23
(The butterfly theorem)
Let PQ be a chord of a circle and
M
be its
midpoint
.
Let AB and CD be two other chords of the circle which pass through the
point
M
.
Let
AD
and BC intersect the chord PQ at the points
X
and
Y
,
respectively
.
Then
MX
=
MY
.
Theorem 4.24
(Maclaurin)
Consider the angle xOy
.
Two points
A
and
B
are mov-
ing on its sides Ox and Oy
,
respectively
,
in such a way that
m
OA
+
n
OB
=
k,
where
m,n
are given positive real numbers and
k
is a given straight line segment
.
Then the circumscribed circle of the triangle OAB passes through a fixed point
.
Remark 4.1
This point lies on the straight line
Oh
which is the geometrical locus
of the points satisfying the property that their distances from the sides of the angle
xOy
are
m/n
.
Theorem 4.25
(Pappou–Clairaut)
Let ABC be a triangle
.
In the exterior of ABC
,
we consider the parallelograms ABB
A
,
ACC
A
,
P
is the common point of the
straight lines
B A
,
C A
,
and
T
is the intersection point of AP
,
BC
.
On the extension of the straight line segment AT and in the exterior of the trian-
gle
,
we consider a straight line segment TN
=
PA
.
Let BB
C C
be the parallelogram such that the point
N
belongs to the side
B C
.
Then the equality
S
BB
C C
=
S
AA
B B
+
S
ACC
A
(4.26)
holds true
,
where
S
denotes the enclosed area of the corresponding quadrilateral
.
Observation
By using the Pappou–Clairaut Theorem, one can easily prove the
Pythagorean Theorem. This can be done as follows (see Fig.
4.12
). We observe
that the right triangles
HTA
and
ABC
are equal. Indeed, it holds
AC
=
AH
and
HT
=
AE
=
AB
,

76
4
Theorems
Fig. 4.12
Illustration of
Pappou–Clairaut
Theorem
4.25
hence
AT
=
BC
=
BK
,
and therefore,
BAM
=
A
1
=
C.
Thus
A
2
+
B
=
C
+
B
=
π
2
,
where
M
is considered to be a point on the side
BC
which is the intersection point
of the straight lines
BC
and
AT
. It follows that
AT
⊥
BC
,
which implies that
AT
BK
.
Since the squares are also parallelograms, by a direct application of the Pappou–
Clairaut Theorem, we deduce that
(
ABDE
)
+
(
ACZH
)
=
(
BKLC
),
that is,
AB
2
+
AC
2
=
BC
2
.

4
Theorems
77
Theorem 4.26
Let ABC be a triangle and
M
be a point in its interior
.
If
x
,
y
,
z
denote the distances of the point
M
from the sides
BC
,
CA
,
and AB
,
respectively
,
of the triangle then the product
xyz
attains its maximum value when the point
M
is
identified with the barycenter of ABC
.
Theorem 4.27
(Cesáro)
Let the triangle ABC be given
.
Consider two triangles
KLM and
K L M
circumscribed around the given triangle and similar to another
given triangle with their respective homological sides perpendicular to each other
,
that is
,
KL
⊥
K L ,
KM
⊥
K M ,
ML
⊥
M L .
Then the sum of the areas of the triangles KLM and
K L M
,
that is
,
S
KLM
+
S
K L M
,
is constant
,
where
S
denotes the enclosed area of the corresponding triangle
.

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