therefore also EA, AB are in one straight line; [1. 14.
join BD.
Then, because the triangle ABC is equal to the trian-
gle ADE, [Hypothesis.
and that ABD is another triangle,
therefore the triangle ABC is to the triangle ABD as the
triangle ADE is to the triangle ABD. [V. 7.
But the triangle ABC is to the triangle ABD as the base
CA is to the base AD, [VI. 1.
and the triangle ADE is to the triangle ABD as the base
EA is to the base AB ; [VI. 1.
therefore CA is to AD as EA is to AB. [V. 11.
Next, let the angle BAC be equal to the angle DAE, and let the sides about the equal angles be reciprocally proportional, namely, CA to AD as EA is to AB: the triangle ABC shall be equal to the triangle ADE.
For, let the same construction be made.
Then, because CA is to AD' 'as EA' 'is to AB, [Hypothesis.
and that CA is to AD as the triangle ABC is to the
triangle ABD, [VI. 1.
and that EA is to AB as the triangle ADE is to the
triangle ABD, [VI. 1.
therefore the triangle ABC is to the triangle ABD as the
triangle ADE is to the triangle ABD ; [V. 11.
therefore the triangle ABC is equal to the triangle ABD [V. 9.
Wherefore, equal triangles &c. q.e.d.
PROPOSITION 16. THEOREM.
If four straight lines he proportionals, the rectangle contained by the extremes is equal to the rectangle con-tained by the means; and if the rectangle contained by the' extremes he equal to the rectangle contained by the means, the four straight lines are proportionals.
