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 DE 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; therefore BL, CM 
are equal and parallel. 
Therefore BLMC is a parallelogram; and its angle LBK is 
equal to the sum of the angles ABC, DHA. 
Now O ABLE = □ BLHA, in the same parallels, 
= O BLNK, for the same reason. 
Similarly □ ACFG = □ AC Mil = o NKGM. 
Therefore, by addition, □ ABLE + □ ACFG — □ BLMC. 
It has been observed (by Professor Cook Wilson x ) that the 
parallelograms on AB, AG need not necessarily be erected 
outwards from AB, AG. If one of them, e. g. that on AG, 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, 
n BLMC = □ BLNK - □ GMNK 
= o ABDE— □ ACFG. 
Again, if both ABDE 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). 
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