Page:Carroll - Euclid and His Modern Rivals.djvu/90

52 Euc. You must insert 'or BE,' to make the comparison fair.

Min. Certainly. I will mark the necessary insertions by parentheses. 'Through C draw CF parallel to AD (or BE), meeting DE in F.'

Euc. You may omit those four words, as they do not occur in my proof.

Min. Very well. 'Cut off (from CF) CG&thinsp;=&thinsp;CB. Through G draw HK parallel to AB (or DE). It is easily shewn that CK, HF are squares of CB, AC; and that AG, GE, are each of them rectangle of AC, CB.'

Euc. We can't admit 'it is easily shewn'! He is bound to give the proof.

Min. I will do it for him as briefly as I can. '∵ CG&thinsp;=&thinsp;CB, and BK&thinsp;=&thinsp;CG, and GK&thinsp;=&thinsp;CB, ∴ CK is equilateral. It is also rectangular, since angle CBK is right. ∴ CK is a Square.' I'm afraid I mustn't say 'Similarly HF is a Square'?

Euc. Certainly not: it requires a different proof.

Min. 'Because CE&thinsp;=&thinsp;AD&thinsp;=&thinsp;AB, and CG, CB, parts of them, are equal, ∴ remainder GF&thinsp;=&thinsp;remainder AC, &thinsp;=&thinsp;HG. But HD&thinsp;=&thinsp;GF, and DF&thinsp;=&thinsp;HG; ∴ HF is equilateral. It is also rectangular, since angle HDF is right. ∴ HF is a Square, and&thinsp;=&thinsp;square of AC. Also AG is rectangle of AC, CB.' I fear I can't assume GE to be equal to AG?

Euc. I fear I cannot permit you to assume the truth of my I. 43.

Min. 'Also GE is rectangle of AC, CB, since GF&thinsp;=&thinsp;AC, and GK&thinsp;=&thinsp;CB. ∴ AG and GE&thinsp;=&thinsp;twice rectangle of AC, CB.'