Pl.z.F.5.

Б

105. COROLL. 3. Hence Parallelograms having equal Bafes and equal Altitudes, are equal.

106. THEOREM 24. If a Triangle and Parallelogram ftand upon the fame Base, and are between the fame Parallels, the Triangle is half of the Parallelogram.

For let BDC be the A, and let AB bell and = a 97. DC, and AD || and = BC, then is ABCD a * Parallelogram, on the fame Bafe and betwixt the fame Parallels. Now, one Side BD of the ABDC is a Diagonal to the Parallelogram ABCD, and the A BDC the Parallelogram ABCD: But the 102. Parallelogram ABCD is any other Parallelogram upon the fame Bafe and between the fame Parallels, ..the ABDC of any Parallelogram on the fame Base and between the fame Parallels. Q E. D.

45.

I

2

=

107. COROLLARY I. Hence, if a Parallelogram and Triangle have equal Bafes, and are between the fame Parallels, or, which is the fame Thing, bave 99. equal Altitudes, the Triangle is Half of the Parallelo

gram.

a

108. COROLL. 2. Hence, if a Triangle and Rectangle bave equal Bases and Altitudes, the Triangle is Half the Rectangle.

109. COROLL. 3. Hence, if two Triangles have equal Bafes and Altitudes, they are equal to each other.

For the Triangles being the Halves of Parallelograms of equal Bafes and Altitudes, which Parallelograms are equal, it follows, that if the Wholes (viz. the Parallelograms) are equal, their Halves (viz. the As) must be equal also.

10. SCHOLIUMI. On thefe two laft Theorems and their Corollaries are grounded the Methods of meafuring, or finding the Areas, or fuperficial Contents, of Parallelograms and Triangles. But in Order to this it must be first fhewn how to find the Area of a Rectangle, which may be thus illuftrated. Let AB Pl.z.F.ro. be a Line which we fuppofe to be moved towards DC, fo as to continue always || to its firft Pofition, viz. let the Point A be moved in the Line AD, and the Point B in the L Line BC; then it is manifeft by fuch Motion it will defcribe a Rectangle. And if we imagine the Line AB to be divided into any Number of equal Parts, when the Line is gone in Length one fuch Part towards D, viz. got into the Pofition ab, it will have defcribed a Rectangle AabB, containing as many little Squares as AB was divided into Parts; and when it gets into the Pofition de, two fuch Parts diftant from its firft Pofition, it will have defcribed a Rectangle containing twice as many Squares as AB contains Parts: And fo when it arrives at DC, it will have defcribed the whole Rectangle ADCB as many little Squares (each Side one Part of AB) as there are Units in the Product of the Number of Parts in AB multiplied by the Number of Parts in BC: And this is called Measuring, or finding the Area of fuch Figure. Hence appears the Reafon why the Product of any two Numbers is by Geometricians called the Rectangle.

SCHOLIUM 2. Hence may be feen the Reafon why 3 x7 = 7 × 3, or generally AxB=Bx A; which is demonftrated algebraically in our Mathematical Effays.

111. Hence, fince any Parallelogram is equal to a Rectangle of equal Bafe and Altitude, to measure any Parallelogram we have this Rule. Multiply the Number of Parts into which the Bafe is fuppofed to be

E

divided,

Pl.z.F.11.

a

divided, by the Number of like Parts contained in its Altitude, and the Product will be the Area required : Thus, for Example. If the Bafe of a Parallelogram be 6 Feet, and Altitude 2 Feet, then its Area is = 6x212 fquare Feet.

112. As to Triangles, every Triangle being half of a Parallelogram of equal Bafe and Altitude, Nothing further need be faid.

113. THEOREM 25. The Complements of Parallelograms which are about the Diagonal of any Parallelogram, are equal to each other.

Let ABCD be a Parallelogram whose Diagonal is AC; and EH, FG, the Parallelograms about AC, that is, thro which AC paffes; and BK, KD, the other Parallelograms, which make up the whole Figure, and are therefore called Complements. Then will the Complement BK be the Complement KD. For, because ABCD is a Parallelogram, and AC 97. its Diagonal, the, AABC' AADC; and becaufe EKHA is a Parallelogram whofe Diagonal is AK, the A AEK = ▲ AHK; and for the fame Reafon AKGC AKFC; . if from the AABC we take away the As AEK and KGC, and from the AADC take away the As AHK and KFC, the remaining Quantities must be equal, viz. the Parallelogram BK = the Parallelogram KD. Q. E. D.

Pl.2.F.12.

114. THEOREM 26. The Diagonals AC, DB, of any Parallelogram bifect each other.

For let the Point of their Interfection be denoted 297. by E. Then, because AD is, and [] BC, and 72. the alternate sequal, viz. LACB=¿ DAC,

and CBDADB, the As AED, BEC, have two LS DAE, ADE, and one Side AD of one two Ls ECB, EBC, and one Side BC of the other, 94. each to each, the As AED, BEC, are equal

in all Refpects, and confequently the Side AE = Side EC, and Side DE Side BE, that is, the Diagonals divide each other into two equal Parts. Q. E. D.

115. THEOREM 27. In any right angled Triangle the Square upon the Side, fubtending the right Angle, is equal to both the Squares upon the Sides containing the right Angle.

b

Pl. 2.F.13.

a

65.

34.

Let ABC be the A. If we fuppofe the Line CA produced to G, fince ▲ CAB is a Rt. 4, the BAG will be fo too, and ·.· one Side of the Square upon BA will fall on the Line AG, and fo with the Line CA will make one Line CG. For the fame Reason one Side of the Square on CA will with BA make one Line BH. Now, fuppofe the Squares described, and the Points F, C; B, K; A, D ; A, E; ь joined by Lines; alfo AL || BD. Then because the 4 DBC4 FBA, each being a right one, ··· adding the common ABC, we have the whole DBA the whole FBC; and the Side FB ¢ 46. BA, and BD = DC: Hence the As FBC, DBA, have two Sides of one, FB, BC, and contained FBC two Sides BA, BD, and contained ▲DBA of the other, each to each, and As FBC, DBA are each other in all Refpects: But FB, GC, are by the Nature of Squares (for the FBA= GAB, being each a Rt. 2, and. FB || GC ;)

f

d

& 58.

e

40.

the Square BG and AFBC are on the fame Base, and between the fame Paraliels, and the Square BG twice the AFBC; and BD, AL, being 106. Parallels, the Parallelogram BL twice the ADBA; but the As FBC, DBA, have been juft fhewn to be = each other, confequently the Square BG Pa- & 45. rallelogram BL. After the fame Manner it may be fhewn that the Square CH the Parallelogram CL;

the Square BG + the Square CH (the Pa- 46. rallelogram BL + the Parallelogram CL, that is =) the Square BE. Q. E. D.

116. SCHOLIUM. This is one of the most extenfive and valuable Theorems in Geometry. - For one Example of its Ufe fee 1ft. of the Mathematical Effays, Article 462.We fhall give another, and eafler Demonftration after we have treated of fimilar Triangles; when it will also be fhewn to be only a particular Cafe of a more general Theorem.

117. Def. 1.

BOOK II.

DEFINITIONS.

Very right-angled Parallelogram is faid to be contained by any two of the Lines which contain one of the right Angles.

118. Def. 2. In every Parallelogram either of the Parallelograms about the Diameter, together with Pl.z.F.11. the two Complements, is called the Gnomen. Thus the Parallelogram EH, together with the Complements BK, KD, is the Gnomon, which is more compendiously expreffed by the Letters BHF, or GED, which are at the oppofite Angles of the Parallelograms which make the Gnomon.

Pl.2.F.14.

119. THEOREM 1. If a Line AB be divided into any two Parts in C, the Square of the whole Line AB, viz. ABED, is equal to the Squares of the two Parts AC, CB, together with twice the Rectangle contained by the Parts AÇ, CB.*

*

Let

It may be demonftrated algebraically thus. Let r = AC, x= CB, then + x = AB = AD, and . the Area of the Square ABED* =r + xxr+ x = (by 1ft. of the Mathemat. Elays, Art. 455.) r2 + 2rx + x2, which is the fame as the Theorem. QE. D. Hence, if xr, by writing for x=rr +2rrrr4rr, the Corollary.