A history of Greek mathematics - Wilbourhall.org

370 PAPPUS OF ALEXANDRIA
contained by BC, HA in an angle which is equal to the sum of
the angles ABC, DHA!
Produce HA to meet BC in K, draw BL, CM parallel to KH
meeting BE in L and FG in M, and join LNM.
Then BLHA is a parallelogram, and HA is equal and
parallel to BL.
Similarly HA, CM are equal and parallel ;
are equal and parallel.
therefore BL, CM
Therefore BLMC is a parallelogram ; and
equal to the sum of the angles ABC, DHA.
its angle LBK is
Now a ABBE — BLHA, in the same parallels,
= O BLNK, for the same reason.
Similarly ACFG = O jiOAffl' = TOCif.
Therefore, by addition, ABDE+C3 ACFG = a 5ZM7.
It has been observed (by Professor Cook Wilson *)
that the
parallelograms on A B, AC need not necessarily be erected
outwards from AB, AC. If one of them, e.g. that on AC, be
drawn inwards, as in the second figure above, and Pappus's
construction be made, we have a similar result with a negative
sign, namely,
o BLMC = BLNK - o CMTO
Again, if both ABBE and ACFG were drawn inwards, their
sum would be equal to BLMC drawn outwards. Generally, if
the areas of the parallelograms described outwards are regarded
as of opposite sign to those of parallelograms drawn inwards,
1
Mathematical Gazette, vii, p. 107 (May 1913).