IF a Parallelogram have the same Base with a Triangle, and be between the same Parallels, then is the Parallelogram double to the Triangle.
If the Parallelagram ABCD, and
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IF a Parallelogram have the same Base with a Triangle, and be between the same Parallels, then is the Parallelogram double to the Triangle.
If the Parallelagram ABCD, and
the Triangle EBC have the same Base BC, and between the same Parallels AE, BC; the Parallelogram shall be double to the Triangle, draw the Line AC.
Demonstration. The Triangles ABC, BCE, are equal (by the 28th.) Now the Parallelogram ABCD is double to the Triangle ABC, (by the 34th.) it is therefore double to the Triangle BCE.
* 1.1 THe general method of measuring of the Area or Superficies of a Triangle, is grounded on this Proposition. Let the Triangle ABC be proposed, we draw from the Angle A the Line AD Per∣pendicular to the Base BC; and Multi∣plping the Perpendicular AD by the Semi-Base BE, the Product gives the Area of the Triangle; because Multi∣plying AD or EF by BE, we have the Content of the Rectangle BEFH, which is equal to the Triangle ABC. For the Triangle ABC is half of the Rectangle HBCG, (by the 41st.) as well as BEFH.
* 1.2 We measure all sorts of Right Lined Figures, as ABCDE, Reducing them first into Triangles BCD, ABD, AED; by drawing the Lines AD, BD, and the Perpendiculars CG, BF, EI. For Multiplying the half of BD by CG, and the half of AD by BF, and by EI, we have the Area of all those Triangles; which added together is equal to the content of the Right Lined Figur•• ABCDE.
* 1.3 We find the Area of Regular Polygons, by Multiplying one half of their peripheries by the Perpendicular drawn from their Centers to one of their Sides; for Multi∣plying IG by AG, we have the Rectangle HKLM equal to the Triangle AIB: And doing the same for every Triangle, taking always the Semi Base, we have the Rectangle HKON, whose Side KO Composed of the Semi-Bases, and conse∣quently equal to the Semi-Periphery; and the Side HK equal to the Perpendicular IG.
According to this Principle, Archi∣medes hath Demonstrated that a Circle is equal to a Rectangle comprehended under the Semi-Diameter, and a Line equal to the Semi-Circumference.
Use 39.
Use 39.
Use 39.