WOLFRAM|DEMONSTRATIONS PROJECT

Triangles with Equal Area Are Equidecomposable (Equivalent by Dissection)

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construct triangle
1.2
l, parallel to base DE
circle with center D
and radius AC
triangle DEG
A dissection of a polygon
X
is a set of polygons that cover
X
exactly without gaps or overlaps. Two polygons are equidecomposable (or equivalent by dissection) if there is a dissection of one that can be reassembled to form the other. In 2D, two polygons of equal area are equidecomposable. On the other hand in 3D, even the cube and the tetrahedron of equal volume are not equidecomposable.
The proof of the 2D case depends on the equidecomposability of two triangles of equal area, which can be split into two parts. The first part is a proof that two triangles with equal bases and equal altitudes are equivalent by dissection. This was proved in the Demonstration Two Equidecomposable Triangles.
It remains to prove that any two triangles of equal area are equidecomposable. So suppose that the two triangles
ABC
and
DEF
have the same area
α
but unequal bases, and that
AB>DF
. The aim is to construct a triangle
DEG
, again of the same area
α
, such that
AB=DG
, so that the previous case can be applied.
Start with a triangle
DEF
with the same area
α
. (For example, if
C
is at height
h
from
AB
, choose
DE
arbitrarily and take
F
at height
hAB/DE
.) Suppose
AB>DF
. Draw a parallel
l
to
DE
through
F
. Let
G
be the intersection of
l
and the circle of radius
AB
with center
D
. All three triangles
ABC
,
DEF
and
DEG
have the same area. The triangles
DEF
and
DEG
have the same base
DE
and equal altitudes, so they are equidecomposable, say using a set of polygons
P
. The triangles
ABC
and
DEG
have equal bases
AB
and
DG
, so they are equidecomposable, say using a set of polygons
Q
. Then
ABC
and
DEF
are equidecomposable using the intersections of
P
and
Q
in
DEG
.