The Algebra Symposium: Geometric Delicacies Discussion
Best for printing:
http://www2.math.uic.edu/~jlewis/as/asrhombusdiscuss.pdf
From: Bettina Pedemonte and Elisabetta Robotti, Aspetti linguistici
della dimostrazione (Linguistic Aspects of Proof), Notiziario
Unione Matematica Italiana, XXXI, No. 10, October, 2004,
pp. 12-30.
2.
ABC is a arbitrary triangle. On the exterior of each side
of the triangle, a square is constructed. Joining the free corners
of each square, three new triangles are created.
Compare the area of each of the three new triangles with the area
of the triangle ABC.
Discussion: Put some additional labels on the
figure.
Compare the areas of ∆ABC and ∆GCI.
Area(∆ABC)
=
1
2
b h
=
1
2
CB ·AP,
Area(∆GCI)
=
1
2
IC ·GQ.
Note that CB = IC, and ∆GQC ≅ ∆APC so that AP = GQ. Hence
Area(∆ABC)
= Area(∆GCI).
More...
One might ask if the argument works if ∠ACB is obtuse
(here ≈ 117°).
Yes!
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