Graphics Reference
In-Depth Information
1
2
a
m
B
B
A
′
c
1
2
a
B
′
m
C
m
A
A
C
1
2
b
1
2
b
Fig. 11.7.
Balancing the triangle along
BB
.
C
m
C
1
2
b
1
2
a
B
′
A
′
P
1
2
b
1
2
a
′
m
A
+
m
B
C
A
m
A
m
B
B
1
2
c
1
2
c
P
is the centroid of the triangle.
Fig. 11.8.
1
2
b
Once more, the triangle will balance, because
m
C
=
m
A
and
B
will be
from
C
and
1
2
b
from
A
.
3. Finally, we do the same for
C
and the edge
AB
. Figure 11.8 shows the final
scenario.
Ceva's Theorem confirms that the medians
AA
,
BB
and
CC
are con-
current at
P
, because
1
1
1
AC
C
B
·
BA
A
C
·
CB
B
A
2
c
1
2
a
1
2
b
1
2
b
=1
Arbitrarily, we select the median
C
C
.At
C
we have an effective mass of
m
A
+
m
B
and
m
C
at
C
. For a balance condition
=
2
c
·
2
a
·
C
P
=
m
C
×
PC
(
m
A
+
m
B
)
×
1
3
and as the masses are equal,
C
P
must be
along the median
C
C
.
And if we use (11.8) we obtain
P
=
1
3
A
+
1
3
B
+
1
3
C
which locates the coordinates of the centroid correctly.
