B

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THEOREM XVIII.

Containing the 35th, 36th, 37th, and 38th of Euclid. Parallelograms, or Triangles, having the fame or equal Bafes, and equal Altitudes, are equal.

The Parallelogram ABCD is equal to the Parallelogram AFGD; standing on the fame Base A D, and between the fame Parallels AH, BG; i. e. they have the fame.or equal Altitudes.

Alfo, the Par. ABCD is equal to
EFGH, having equal Bafes, AD EH.

DEM. Now, AB-CD, and AF=DG.
And, the Angle BAF CDG.

Wherefore, the Triangle ABFDCG.

Th. 15.

6.

8.

Ax. 7.

But, the Triangle ICF is common to both;
wherefore, the Trapezium ABCI IFGD.
Add, to both, the Triangle AID; which is common;
the Parallelogram ABCD is equal to AFGD - Ax. 6.

Again. The Par. ABCD, is proved equal to AFGD.
For the fame reason, EFGH-AFGD, FG being common.
Therefore, the Par. ABCDEFGH.
Ax. 3.
But, AD FG and EHFG (15); wh. AD=EH.-Ax.3.
Th. the Parallelograms, AC and FH, have equal Bases.

Secondly. Draw the Diagonals BD, DF and EG.
Now, fince every Triangle is equal to half a Parallelogram
Th. 17.
having the fame Base and Altitude,
the Tri. ABD half the Par. ABCD; &, AFD-half AFGD
Therefore, the Triangle ABD is equal to AFD. - - Ax. 4.

And, for the fame reason, the Triangle ABD=EGH. (For, the Parallelogram ABCD is equal to EFGH.) And, their Bases, AD, EH, are also equal.

Q. E. D.

COR.

COR. Equal Parallelograms or Triangles, having the fame or equal Bafes, have the fame Altitude; or, are contained between the fame Parallels.

This Corollary is the converfe of the Theorem, and contains the 39th and 40th Propofitions of Euclid.

N. B. From this Theorem, the 24th, 26th, and 27th Problems are deduced; which may be found particularly ufeful to Surveyors of Land, &c.; feeing, by its means, any Plot or Survey, though ever fo irregular, in Figure, may be reduced, with the greatest facility, and accuracy, to a Trapezium or Triangle; and, its Area obtained at one Operation. (See Prob. 24.)

Alfo, by its means, almoft any right-lined Figure, as well as a Trapezium, may be divided into two equal Parts, by a Right Line from any detetmined Point, in any Side.

An Example, of which, will be found in the Appendix.

SCHOL. Hence it is evident, that two Spaces may differ greatly in Circumference, yet contain the fame Space; and also, that two Figures or Spaces, of equal Circuit, may contain very different Areas.

In

THEOREM XIX. 43. Euclid.

every Parallelogram, the Complements are equal.

In the Parallelogram ABCD; affume any point, as I, in the Diameter AC; through which, draw EF and GH, parallel to AB and AD.

I fay, the Complements, DI and IB, are equal to each other.

E

DEM. AC divides the Par. DB into two Tris. ABC=ACD. And the Parallelograms AGIE and IFCH, are divided into equal Triangles AGI=AIE, and IFC=CH. Th. 15. Now, if from the equal Triangles ABC, ACD, there be taken away the equal Triangles AGI, IFC=AIE, ICH, there will remain, the Par. DEIH equal to IGBF. - Ax 7. TH E O.

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THEOREM

XX.

47. Euclid.

In every Right-angled Triangle, the Square of the
Hypothenufe, is equal to the Squares, of the two
Sides, containing the Right Angle.

K

Let ABC be a Right-angled Triangle.

On the Hypothenufe AC, defcribe the Square ACDE; and, on the Sides AB, CD, describe the Squares AFGB and BHIC.

I fay, the Square AD, of the Hyp. AC is equal to the two Squares, AG and BI, of the Sides, AB and BC.

From the Right Angle, B, draw BK
perpendicular to AC; which, produce to L, parallel to CD.
The Square AG is equal to the Rectangle AL; and, the
Square BI is equal to the Rectangle KD. Draw FC & BE.

DEM. In the Triangles FCA, ABE, AF, AC are respec-
tively equal to AB, AE;. and they contain equal Angles.
For, FAB EAC (Ax. 9.) and, if BAC be added to both,
the Angle FAC=BAE.
Ax. 6.
Wh. FC BE, and the Triangle FCA ABE. Th. 8.

But, the Tri. FCA is equal to half the Square FB - 17. (for, they have the fame Bafe, FA; and, they are between the fame Parallels, GC and FA.)

And, the Tri. ABE (eq. FCA)=half the Rect. AKLE; (on the fame Bafe, AE, & between the Parallels BL & AE) Therefore, fince the Tri. FCA ABE (and they are each equal to half the Square AG, or the Rectangle AL) the Square AFGB is equal to the Rectangle AKLE - Ax.5.

Again; the Square BI is equal to the Rectangle CKLD;
Draw Al and BD.

Then, in the Triangles AIC, CBD, the Sides IC,
CA, are equal to BC, CD, refpectively.

And, the Angle ICA BCD (ACB being common)-Ax.6. wherefore, the Triangle AIC=CBD. - - Th. 8.

But, the Triangle IAC, half the Square CH; - 17. (having the fame Bafe, CI, and between the Pars.AH, CI) And the Tri. CBD (eq. IAC)= half the Rest. KD-17. (being on the fame Bafe, DC, & between the Pars. LB, DC) Wh. the Square BI is equal to the Rectangle KD-Ax. 5. But, the Square AG was proved equal to the Rect. AL. and AL+ KD AD (the Square of the Hyp. AB.)-Ax. 2. Therefore, the Square AFGB+BHIC=ACDE. QE D.

Or, it may be, as elegantly and more briefly, demonftrated after the following manner.

Then, the Square AG is equal to the Rectangle AL; and the Square CH is equal to the Rectangle CL.

DEM. For, the Triangle AEB half the Square AFGB -17.
Having the fame Bafe, AB, and Altitude, BG.

The Tri. AEB is alfo equal half the Rect.AELK - fame
having the fame Bafe, AE, and Altitude, AK.
Th. the Square AG is equal to the Rect. AL. - - Ax. 5.

K

Again; the Tri. BDC half the Square BHIC; - 17. (on the fame Bafe, BC, & between the Parallels DI, BC), And the Tri. BDC is also equal half the Rect. KLDC; (having the fame Bafe, CD, and Altitude, LD.) Wh. the Square BHIC = the Rect. KLDC. - Ax. 5. But, the Rect. AL+LC= the Square AEDC. - Ax. 2. Th. the Square AEDC (of the Hyp. AC) is equal to the Squares, AFGB and BHIC, of the two Sides, AB & BC.

COR. Hence; if, in a Triangle, the Square of one Side be equal to the Squares of the other two Sides, the Angle, contained by thofe two, is a Right Angle.

I have been more particular in the Demonftration of this famous Theorem, than any of the preceding ones; not because it is the laft of this Book, but, because it is fo extraordinary in itself, and of fuch fingular ufe throughout the Mathematics. It is alfo of great ufe in the following Books of the Elements; infomuch, that, without it, we fhould not be able to advance much further. Pythagoras is faid to be the inventor of it; for joy of which, it is faid, that he offered a hundred Oxen to the Deity, who infpired him with fo noble, and fo ufeful, a Theorem.

That there may not remain the least doubt, in the mind of any, concerning the truth it contains, I have added another Figure; the Demonftration of which is both ocular and mathematical.

EX. 3. Let the Square AEDC, of the Hypothenufe AC, be inverted, as before; and, having drawn the Squares of the two Sides, AB, and BC; produce FA and FG, IC and IH, meeting in K and L; forming the Square FLIK.

The two Rectangles ABCK, and BGLH, are equal; being under equal Lines, AB=BG, and BC=BH. And the four Triangles ACK, AFE, ELD, and DIC, are equal between themselves, and alfo to the Rectangles, BL, BK; each being equal to half a Rectangle; as ACK, equal half ABCK.

Th. 15

But,