Heron Alexandrinus
,
Mechanica
,
1999
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[8]
As
for
the
case
that
a
point
that
is
moving
in
two
motions
of
constant
speed
can
follow
unequal
lines
,
we
are
going
to
prove
that
now
.
Let
us
assume
a
rectangle
,
namely
<abgd>,
and
let
the
line
<ad>
be
a
diagonal
;
further
let
the
point
<
a
>
run
in
steady
motion
on
the
line
<
ab
>
and
let
the
line
move
in
steady
motion
on
the
two
lines
<ag>, <bd>,
so
that
it
is
always
parallel
to
line
<gd>;
also
let
the
time
in
which
point
<
a
>
runs
to
<
b
>
be
equal
to
the
time
in
which
line
<
ab
>
reaches
<gd>;
thus
I
say
that
point
<
a
>
in
a
certain
time
covers
two
unequal
lines
.
Proof
for
this
:
If
line
<
ab
>
moves
for
a
certain
time
and
is
placed
on
line
<ez>,
then
the
point
moving
on
line
<
ab
>
in
the
same
time
comes
to
rest
on
line
<ez>
and
a
constant
relation
sets
in
.
The
relation
between
line
<ag>
and
line
<
ab
>,
i
.
e
.,
with
line
<gd>,
is
,
namely
,
equal
to
the
relation
of
line
<
ae
>
with
the
line
that
lies
between
point
<
e
>
and
the
point
moving
on
it
.
Line
<ag>,
however
,
is
to
line
<gd>
as
<
ae
>
is
to
<
eh
>.
Thus
the
point
moving
on
line
<
ab
>
after
<
h
>
falls
on
the
line
<ad>,
which
is
the
diagonal
.
Similarly
we
prove
that
the
point
running
on
line
<
ab
>
constantly
moves
forward
on
line
<ad>
and
is
at
the
same
time
moving
on
lines
<ad>
and
<
ab
>.
The
two
lines
<ad>
and
<
ab
>
are
,
however
,
different
;
thus
,
the
point
moving
forward
in
steady
motion
covers
unequal
lines
in
the
same
time
.
But
,
as
I
said
,
the
motion
of
the
point
on
line
<
ab
>
is
simple
,
its
motion
on
the
diagonal
<ad>
however
is
derived
from
the
motion
of
<
ab
>
on
the
two
lines
<ag>
and
<bd>
and
the
motion
of
<
a
>
on
line
<
ab
>.
Thus
the
one
point
<
a
>
covers
in
steady
motion
two
unequal
lines
.
q
.
e
.
d
.
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