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Similarly, we can state that
−
DE
−
DB
=
(1.5)
where is some scalar.
We now want to prove that
1
2
1
2
. We begin by rewriting (1.4) substituting (1.2):
=
and
=
−
AE
=
r
+
s
=
r
+
s
(1.6)
We can also rewrite (1.5) substituting (1.3):
−
DE
=
r
−
s
=
r
−
s
(1.7)
So, now we know
−
AE and
−
DE. But we want a third equation that associates
−
AE with
−
DE. Looking
at Fig. 1.13, we see that in the triangle AED we have
−
AE
=
−
AD
+
−
DE
+
−
DE
=
s
(1.8)
Therefore, substituting Eqs. (1.6) and (1.7) in Eq. (1.8), we obtain
r
+
s
=
s
+
r
−
s
and
r
+
s
=
r
+
1
−
s
Now we have an equation linking , ,
r
, and
s
. But as
r
and
s
are non-collinear, we can equate
coefficients: therefore,
=
and
=
1
−
which means that
=
1
−
and
1
2
=
Also,
1
2
=
Therefore, the diagonals of a parallelogram bisect each other.
The time has now come to formalize these concepts and see how vectors can be encoded
using Cartesian coordinates.
