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But the Trapezium A B C + ACxD=D ABCD and the Trapezium ab x D + a CxD=abCD, consequently, BABCD=26CD. Q. E. D.

Eorollary.

Hence it will be easy to conceive, that all Triangles which stand upon the fame Base, or upon equal Bases, and between the same Parallels, (viz. having the fame Height) are equal one to another. (37 & 38 e. 1.)

For all Triangles are the Halfs of their circumscribing Parallelograms; and therefore, if the Wholes be equal, their Halfs will also be equal.

THEOREM XXI.

Parallelograms (and consequently Triangles) which have the same

Height, have the same Proportion one to another as their Bases have. (1. 6, 6.)

Demonttration.

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Like Triangles are in a duplicate Ratio to that of their homologous

Sides. (19. 6. 6.) That is, the Area's of like Triangles are in Proportion one to another as are the Squares of their like Sides.

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Let A and a be Perpendiculars to the two Bases D and d. Then DA= the Area of ABCD

}By Lemma 3, Page 303. And da= the Area of Abc d

But 1 | B:6::D:d}&c. By Theorem 13.
And B:6:: 4:a

c Conseq. / 3]D:d:: 4:a 3:14

Da=dA 4X;D 5DDda = |Dddd. By Axiom 3. 5 Hence 16 DD:dd::DA:{da. And so for other Sides.

Q. E. D.

2

THEOREM XXIII.

In every Obtuse-angled Triangle (as B C D) the Square of the Side

fubtending the obtuse Angle (as D) is greater than the Squares of the other two Sides (B and C) by a double Rectangle made out of one of the Sides fas B) and the Segment or Part of that Side produced, (as a) until it meet with the Perpendicular (P) let fall upor it. (12. 6. 2.) That is, DD= BB+CC + 2 B a.

Demontration. First 1| DD=PP taa+2Ba+BB

And 2 CC=PP-taa 1 - 213 | DD-CC=2Ba+BB itcc 4 | DD=BB+CC+2Ba

D

B

a

Q. E. D. Corcllary. Hence 'tis evident, that, if the sides of any Obtuse-angled Triangle are given, the Segment (a) of the Side produced (or the Perpendicular, P) may be easily found.

THEOREM XXIV.

If a Perpendicular (as P) be let fall in any Acute angled Triangle

(as B C D), the Square of either of the Two Sides (as D) is less than the Squares of the other Side, and that Side upon which the Perpendicular falls (viz. C and B) by a double Rectangle made of the Side B, and that Segment or Part of it (viz. a) which lies next to the Side C. (13. 6. 2.) That is, DD + 2 Ba = BB + cc.

Demonto

J

aa

Demonttration.
Firft|1| DD=PP+" } By Theo. 11.
And 2 CC=PP+aa }
But

3
Ba=e, by Figure,

D

с 3 24

P BB - 2Ba + aa =le. - aa 5 BB Ba = leaa.

е 4

a 216 DDCC = e

B 67 DD- CC = BB 2 Ba 7 8|DD + 2 Ba= BB + CC.

Q. E. D. Corollary. Hence it follows, that, if the sides of any Acute-angled Triangle be known, the Perpendicular P, and the Segments of the Side whereon it falls (viz. a, e.) may be easily found.

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CHAP. IV.

The Solution of several Easy Probleins in Plain Geometry;

whereby the Learner may (in Part) perceive the Applica

tion or Use of the foregoing Theorems. N

O TE, wisen a Line, or the side of any plain Triangle, is an

Way cut into two (or more Parts, either by a Perpendicular Line let fall upon it, or otherwise, those Parts are usually calld Segments ; and so much as one of those Parts is longer than the other, is calld the Difference of the Segments.

And when any Side of a Triangle, or any Segment of its Side is given, 'tis usually mark'd with a small Line cross it, thus: and these Sides, or Parts of Sides, that are sought, are mark'd with four Points, thus:

PROBLEM I. To cut or divide a given Right-line (as S) into Extream and

Mean Proportion. (11. e. 2:)
That is, to divide a Line so, that the Square of the greater
Segment (or Part) a, may be equal to the Rectangle made of the
whole Line S, and the lefser Segment e.

S
Viz. 1 1 , Se = na, by the Problemi.
And S -a meg for Sate

2

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315

aa

Sa

2 and

3 4 3=S-á. By Axiom 5. 4 x S

6aat Sa = SS 6, solved

zla=VSSTISS::SS. See Pages 195, 196.

5 aa=SS

5 + Sa

Note, The last Problem cannot be truly answer'd by Numbers, but Geometrically it may be performed, thus :

1. Make a Square, whose Side is = S the given Line, and bisect one of its Sides in the Middle, as at C; upon the Point C describe fuch a Semicircle as will pass through the remoteft Points of the Square, and compleat its Diameter.

a

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2. Then will either Part of the Diameter, on each End of the Side S, be=a, the greater Segment fought.

But a+S:S:S:a, By Theorem 13.
Ergo, a at Sa=SS. Which was to be done.

PROBLEM II.

The Base of any Right-angled Triangle, and the Difference between

the Hypothenuse and Cathetus being given, 'to find the Cathe

tus, &c.

Let {

b = 72

I

2 d = 32
And 312= Čathetus fought
Then
4 bb taa=dd + 2da faa

By Theorem II. 4

5

bb=dd + 2da Š dd 6 | 2da = bb - dd

bb-dd 6 • 2d 7

= 65 Or, 86:d + 2a::d: b. By Theorem 13. 8

9

bb =dd +2da. As before at the 5th Step. .

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Here you see that either Way raises the same Æquation ; neither is their any constant Method or Road to be observ'd in solving Geometrical Problems, but every one makes Use of such Ways and Theorems as happen to come first into their Mind, the Result being every Way the same.

PR O B L E M III.
The Difference between the Base and Hypothenuse of any Right-angled

Triangle, and the Difference between the Cathecus and Hypothe-
nuse being both given, to find the Triangle.
Let {

2X=25
And 3 dtxta=the Hypot.

:e Then

4 dtasy

Id=32

5x+aze} by the Proble

4

6 dd +2da taa=9y 5

7 *x+2xa taa=ee 3 dd2dx+2da+2xa -xxtaa=0 Hypothenus. 6 to 71 91dd+ada+2xa+*x+2aa=yytee.

The two last Steps are equal, by Theorem 11. Consequently, if those Things that are equal in both be taken away, the Remain. ders will be equal. By Axiom 2.

That is aa = 2dx = 1600 10 we? a = V 2dx = 40 If II dta = 72 = y The Base. 3 + 1 13 * ta = = 65 = e The Cathetus. it2tu 14 d + x +a= 97 The Hypothenuse.

IO II 12

PROBLEM IV.

The Hypothenuse, and the Sum of the other trvo Sides, of ang Right

angled Triangle, being given, thence to find the Sides.

1

H=97

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Let

H
And

2

ate=S=137 By Fig. 3 aate=HH 2

4 aa+2aeteer SS 4 3 5 201=SS-HH 3 5 6

2ae + ee=2HH.SE 6

7 a-e-V HH-S5

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