Graphics Reference
In-Depth Information
n
2
n
2
+
n
+1
B
+
1
n
2
+
n
+1
A
+
n
n
2
+
n
+1
C
E
=
n
n
2
+
n
+1
A
+
1
n
2
+
n
+1
B
+
n
2
n
2
+
n
+1
C
F
=
As a quick test for the above equations, let
n
=1,whichshouldmake
D
,
E
and
F
concurrent at the triangle's centroid:
D
=
1
3
A
+
1
3
B
+
1
3
C
E
=
1
3
A
+
1
3
B
+
1
3
C
F
=
1
3
A
+
1
3
B
+
1
3
C
which is rather reassuring.
Now let's return to the final part of the problem and determine the area of
triangle ∆
DEF
in terms of ∆
ABC
. The strategy is to split triangle ∆
ABC
into four triangles: ∆
BCF,
∆
CAD,
∆
ABE
and ∆
DEF
as shown in Figure
11.23.
Therefore
area
∆
ABC
=
area
∆
BCF
+
area
∆
CAD
+
area
∆
ABE
+
area
∆
DEF
and
1=
area
∆
BCF
area
∆
ABC
+
area
∆
CAD
area
∆
ABC
+
area
∆
ABE
area
∆
ABC
+
area
∆
DEF
area
∆
ABC
(11.14)
But we have just discovered that the barycentric coordinates are intimately
connected with the ratios of triangles. For example, if
F
has barycentric co-
C
1
2
B
′
F
2
A
′
E
1
D
C
′
A
1
2
B
Fig. 11.23.
Triangle ∆
ABC
divided into four triangles ∆
ABE
,∆
BCF
,∆
CAD
and
∆
DEF
.
