Make a copy of the hexagon and construct the triangle ACE in one
and the triangle BDF in the other.
In the copy with triangle ACE, construct parallelograms
ABCG, CDEH, and EFAI. In the copy with triangle BDF,
construct parallelograms BCDP, DEFQ, and FABR.
Let [ST..YZ] denote the area of polygon ST..YZ. Then
[ABCDEF] = [ABCG] + [CDEH] + [EFAI] + [GHI]
= [BDCP] + [DEFQ] + [FABR] + [PQR] (1)
Looking at the side lengths of triangles GHI and QRP:
|GH| = | |AB| - |DE| | = |QR|
|HI| = | |CD| - |FA| | = |RP|
|IG| = | |EF| - |BC| | = |PQ|
Therefore, the triangles are congruent and
[GHI] = [PQR] (2)
Combining (1) and (2) gives
[ABCG] + [CDEH] + [EFAI] =
[BDCP] + [DEFQ] + [FABR] (3)
Combining (2) and (3) gives
[ACE] = ( [ABCG] + [CDEH] + [EFAI] )/2 + [GHI] +
= ( [BDCP] + [DEFQ] + [FABR] )/2 + [PQR]
= [BDF]
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