334 11. POST-EUCLIDEAN GEOMETRY
Í
t
X /
w A Ê
A
U
Æ ç Ε Ç
FIGURE 8. Thabit ibn-Qurra's attempted proof of the parallel postulate.
be uncontroversial. The difficult part occurs when Æ is between A and E. Thabit
ibn-Qurra argued as follows. By Archimedes' principle, some multiple of AZ, say
AH, exceeds AE, so that Ε lies between A and H. Now by drawing a perpendicular
to ç at Ç and making HK equal to ZW, we get a Thabit quadrilateral, so that
if WK is joined, we have a rectangle WZHK. Then, if X is chosen so that
AW = WX and a perpendicular XU is drawn to WK, the triangles AW Æ and
WXU will be congruent because the sum of the acute angles of a right triangle is
a right angle. Then WU will equal AZ. We can then start over, since WK will
be less than AH by a length equal to AZ. In this way, in a finite number of steps,
we will reach a point Í on line m that is also on the extension of HK. Hence ôç
contains points on both sides of / and therefore intersects I.
Gray has called Thabit ibn-Qurra's mistake "an interesting and deep one,"
since it makes use of motion in geometry in a way that seems to be implied by Eu-
clid's own arguments involving coinciding figures; that is, that they can be moved
without changing their size or shape. Euclid makes this assumption in Proposi-
tion 4 of Book 1 where he "proves" the side-angle-side criterion for congruence by
superimposing one triangle on another. He does not speak explicitly of moving a
triangle, but how else is one to imagine this superimposition taking place?
Thabit ibn-Qurra also created a generalization of the Pythagorean theorem. His
theorem is easily derived from similar triangles. Consider a triangle ABC whose
longest side is BC. Copy angle Β with A as vertex and AC as one side, extending
the other side to meet BC in point C", then copy angle C with A as vertex and
BA as one side, extending the other side to meet BC in point B', so that angle
AB'B and angle AC'C both equal angle A. It then follows that the triangles B' AB
and CAC are similar to the original triangle ABC, and so AB — BC • BB' and
AC^2 = BC • CC, hence
AB^2 + AC^2 = BC(BB' + CC).
The case when angle A is acute is shown in Fig. 9.
