Prop. If the sum of the interior angles ECF and DBC which two straight lines EC and DB make with a third line CP is less than two right angles, the lines, if sufficiently produced, will intersect.
Proof. Construct PCA equal to the supplement PBD of CBD, and ECF, FCG, etc. each equal to ACE, so that ACF = 2.ACE, ACG = 3.ACE, etc. Then however small the angle ACE may be, there exists some number n such that n.ACE = ACH will be equal to or greater than ACP.
Again, take BI, IL, etc. each equal to CB, and draw IK, LM, etc. parallel to BD, then the figures ACBD, DBIK, KILM, etc. are congruent, and ACIK = 2.ABCD, ACLM = 3.ACBD, etc.
Take ACNO = n.ACBD, n having the same value as in the expression ACH = n.ACE, then ACNO is certainly less than ACP, since ACNO must be increased by ONP to be equal to ACP. It follows that ACNO is also less than ACH, and by taking the nth part of each of these, that ACBD is less than ACE.
But if ACE is greater than ACBD, CE and BD must intersect, for otherwise ACE would be a part of ACBD.
Journal für Mathematik, Bd. 2 (1834), p. 198.
[2112]. Are you sure that it is impossible to trisect the angle by Euclid? I have not to lament a single hour thrown away on the attempt, but fancy that it is rather a tact, a feeling, than a proof, which makes us think that the thing cannot be done. But would Gauss’s inscription of the regular polygon of seventeen sides have seemed, a century ago, much less an impossible thing, by line and circle?—Hamilton, W. R.
Letter to De Morgan (1852).
[2113]. One of the most curious of these cases [geometrical paradoxers] was that of a student, I am not sure but a graduate, of the University of Virginia, who claimed that geometers were in error in assuming that a line had no thickness. He published a school geometry based on his views, which received the endorsement of a well-known New York school official and, on the basis of this, was actually endorsed, or came very near being endorsed, as a text-book in the public schools of New York.—Newcomb, Simon.