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TRIGONOMETRY 
I. To prove I. 28, Ptolemy takes two straight lines AB, CD, 
and a transversal EFGH. We have to prove that, if the sum 
of the angles BFG, FGD is equal to two right angles, the 
straight lines AB, CD are parallel, i.e. non-secant. 
Since AFG is the supplement of BFG, and FGC of FGD, it 
follows that the sum of the angles AFG, FGC is also equal to 
two right angles. 
Now suppose, if possible, that FB, GD, making the sum of 
the angles BFG, FGD equal to two right angles, meet at K; 
then similarly FA, GC making the sum of the angles AFG, 
FGC equal to two right angles must also meet, say at L. 
[Ptolemy would have done better to point out that not 
only are the two sums equal but the angles themselves are 
equal in pairs, i.e. AFG to FGD and FGC to BFG, and we can 
therefore take the triangle KFG and apply it to FG on the other 
side so that the sides FK, GK may lie along GC, FA respec 
tively, in which case GC, FA will meet at the point where 
K falls.] 
Consequently the straight lines LABK, LCDK enclose a 
space : which is impossible. 
It follows that AB, GD cannot meet in either direction; 
they are therefore parallel. 
II. To prove I. 29, Ptolemy takes two parallel lines AB, 
CD and the transversal FG, and argues thus. It is required 
to prove that Z AFG + Z CGF — two right angles. 
For, if the sum is not equal to two right angles, it must be 
either (1) greater or (2) less. 
(1) If it is greater, the sum of the angles on the other side, 
BFG, FGD, which are the supplements of the first pair of 
angles, must be less than two right angles. 
But AF, CG are no more parallel than FB, GD, so that, if 
FG makes one pair of angles AFG, FGC together greater than