But the triangle DAE is equal to the triangle BAC. [Hypothesis.
Therefore the triangle DAE is equal to the triangle DAF. [Ax. 1.
Therefore EF is parallel to AD. [I. 39.
Suppose now that the angle DAE is greater than the angle DAF.
Then the angle CAE is equal to the angle AFE, [I. 29.
and therefore the angle CAE is equal to the angle AFE, [I. 5.
and therefore the angle CAE is equal to the angle BAC. [I. 29.
Therefore the angles BAC and DAE are together equal to two right angles.
Similarly the proposition may he demonstrated if the angle DAE is less than the angle DAF.
VI. 16. This is a particular case of VI. 14.
VI. 17. This is a particular case of VI, 16.
VI. 22. There is a step in the second part of VI. 12 which requires examination. After it has been shewn that the figure SR is equal to the similar and similarly situated figure NH, it is added "therefore PR is equal to GH." In the Greek text reference is here made to a lemma which follows the proposition. The word lemma is occasionally used in mathematics to denote an auxiliary proposition. From the unusual circumstance of a; reference to something following, Simson probably concluded that the lemma could not be Euclid's, and accordingly he takes no notice of it.
The following is the substance of the lemma.
If PR be not equal to GH, one of them must be greater than the other; suppose PR greater than GH.
Then, because SR and NH are similar figures, PR is to PS as GH is to GN. [VI. Definition 1.
But PR is greater than GH, [Hypothesis.
therefore PS is greater than GN. [V. 14.
Therefore the triangle RPS is greater than the triangle HGN. [I. 4, Axiom 9.
But, because SR and NH are similar figures, the triangle RPS is equal to the triangle HGN; [VI. 20.
which is impossible.
Therefore PR is equal to, GH.
VI. 23. In the figure of VI. 23 suppose BD and GE drawn. Then the triangle BCD is to the triangle GCE as the parallelogram ACis to the parallelogram CF. Hence the result may be extended to triangles, and we have the following theorem,
