Now let AB, CD make with AC two interior ∠s BAC, ACD together < 2 right ∠s. It shall be proved that they will meet if produced towards B, D.
In CD take any point D. Join AD. At A make ∠DAH equal to ∠CBA.
Hence AH, CD are equally inclined to all transversals; [(μ).
∴ ∠s HAC, ACD together = 2 right ∠s;
∴ they together > ∠s BAC, ACD;
∴ ∠HAC > ∠BAC, i.e. ∠HAD > ∠BAD;
∴ ∠CDA > ∠BAD.
At D, in Line DA, make an ∠ equal to ∠BAD;
then the Line, so drawn, will fall within ∠CDA, and will meet CA between C and A. Call this point K.
In CA produced take CL a multiple of CK, and > CA. And in CD produced take CM the same multiple of CD that CL is of CK. And join LM.
Because Triangles CKD, CLM have a common vertical ∠, and the 2 sides of the one respectively equimultiples of those of the other,
∴, by what has been already proved, ∠s CKD, CLM are equal.
Because AB, KD are equally inclined to AD,
∴ they are equally inclined to CL; [(μ).
∴ ∠CAB = ∠CKD i.e. = ∠CLM;
∴ AB is separational from LM; [Euc. I. 28.
∴ if produced, it will meet CM.
Therefore a Pair of Lines, &c. Q. E. D.
