The elements of non-Euclidean plane geometry and trigonometry, by H. S. Carslaw.

86, 87] THE RIGHT-ANGLED TRIANGLE

147

Move the triangle bcA along AC till it coincides with C and
be takes up the position b'C.
We thus have the triangle b'a'C
congruent with bAc.
In the same way move the b b
triangle bcA along BA until b
coincides with B and the triangle
takes up the position Ba"c".
Through the middle point I of
a'A draw I L perpendicular to BA.
Then LI produced will be perpendicular to b'a'.
We thus obtain the common
perpendicular to b'a' and BA,
the line KIL.
In the same way we obtain K
the common perpendicular MJN a'/ - s
C C
to AC and a"'c through the j C
middle point J of Aa". C
Finally, we draw b'Q perpen- a" N c
dicular to AB and bb" perpen- FIG. 1(4.
dicular to BC.
We have seen that as Bb tends to zero, we have
bb" b'Q
Lt Bb= Lt............................(i)
MJ IL
In the same way Lt-= Lt-.
JA IA
MN KL
Thus Lt, = Lt..........................(i)
Aa C AAa'
Dividing (i) by (ii) and remembering that Aa"= Bb and
Aa' = Cc, we have
bb" b'Q Cc
Mt MN = KL t B,
which may be written
b'Q bb" Bb'
Lt Lt t I. Lt MB............... (iii)
KL C MN i t s
We shall now show that this equation is the same as

0 (AB)= (BC) (CA).