Discovering and proving geometric inequalities by CAS (pdf)

Figure 3: Parallelogram law: 2(a 2 + b 2 )=e 2 + f 2
Summing up both equalities we get (2).
We proved the statement above both in an automatic and classical way. Both methods
have their strengths and weaknesses. The classical way required basic knowledge from
elementary geometry (we used the law of cosines, etc.), gives us a good insight into the
problem, but had one weakness - we had to have the key idea how to prove the statement.
This is not always easy to find.
On the other hand the automatic proof required basic knowledge from analytical geometry
(”clever” introduction of coordinate system, expression of distances od vertices analytically,
etc.), the computation was quite automatic. But this method was not so beautiful
in geometric sense. This method can also lead to unexpected problems - e.g. to find non
- degeneracy conditions.
When students were asked which method they preferred and why, the answer was: ”the
classical method because of a better insight into the problem.” They agreed with the conclusion
that both methods should be combined and used in practice.
Remark:
The parallelogram law (2) had been already known in a little bit different form to ancient
Greeks. It became familiar since the year 1935, when Jordan and von Neumann showed
that Banach space in which (2) holds, is Hilbert space.
The parallelogram law (2) can be generalized on a trapezoid:
A quadrangle ABCD is a trapezoid with bases a, c, legs b, d and diagonals e, f if and only
if
b 2 + d 2 +2ac = e 2 + f 2 . (3)
We will leave the proof to the reader.
The relation
2(a 2 + b 2 )=e 2 + f 2 .
is a necessary condition for a quadrilateral ABCD to be a parallelogram. Is this condition
also a sufficient condition?
We will prove the following statement:
A quadrilateral ABCD with sides a, b, c, d and diagonals e, f is given. Then ABCD is a
parallelogram if and only if it holds
a 2 + b 2 + c 2 + d 2 = e 2 + f 2 . (4)