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In every Obtufe-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 (as B) and the Segment or Part of that Side produced, (as a) until it meet with the Perpendicular (P) let fall upon it. (12. 2. 2.). That is, DD=BB+CC + 2 Ba.

Demonâration.
Fift | 1 | DD=PP taa+2 Bat BB

And | 2 | CC=PP taa
I-23 | DD-CC=2Ba+BB

D: P itcc 4 | DD=BB+20+2Ba

_ B__.a.

Q. E. D. Corollary. 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 CD), 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 Reflangle made of the Side B, and that Segment or Part of it (viz. a) which lies next pfo the Side C. (13. 6. 2.) That is, DD + 2Ba - BB + CC.

Demons

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Ć HA P. IV. The Solution of several Easy Problems in Plain Geometry,

whereby the Learner may (in Part) perceive the Applica

tion or Use of the foregoing Theorems. N O TE, when a Line, or the side of any plain Triangle, is ang

" Way cut into two (or more Parts, either by a Perpendicular · Line let fall upon it, or otherwise, those Parts are usually caild Sego ments; 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 those Sides, or Parts of Sides, that are fought, 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 Se = aa, by the Problein.
And | 2 | 3 - a é, for $ = a te

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2. Then will either Part of the Diameter, on each End of the Side S, be = 1, the greater Segment sought.

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

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PROBLEM III. The Difference between the Base and Hypothenuse of any Right-angled · Triangle, and the Difference between the Cathetus and Hypothe

nuse being both given, to find the Iriangle.

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PROBLEM IV.
The Hypothenuse, and the Sum of the other two Sides, of any Right-

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

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