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THEOREM I. If a Right-line stand upon (or meet with) another Right-line, and make

Angles with it, they will either be two Right-angles, or two Angles equal to two Right angles (13 e. 1.)

Demontration. Suppose the Lines to be A B and DC, meeting in the Peint at C: Upon C describe any Circle at pleasure ; then will the Arch A D be the Measure of the 5 b, and the Arch D B the Measure of se; but the Arches ADED B=180°, un viz. they compleat the Semicircle. Consequently the sb to se=180°. Which was to be provid.

Corollaries. 1. Hence it follows, that if the 5b=90°then ve=90°; but if sb be obtuse, then the se will be acute, &c.

From hence it will be easy to conceive, that if several Rightlines stand upon, or meet with any Right-line at one and the same Point, and on the same Side, then all the Angles taken together will be = 180°, viz. Two Right-angles.

THEOREM II. If two Angles inter seat (i, e, cut or cross) each other, the two ope

pofite Angles will be equal. (15.4. 1.)

Demonftration.
Let the two Lines be A B and
DE, intersecting each other in the
Center C.
Then <ht Fe=180°
And Sotsa=180° s po
Consequently sbt ser sot.
sa, per, Axiom 5.

Subtract sb on both sides of the Æquation, and it will leave Per ca.

Again, '<bt se=180°, as before ; and set = 180°, confequently set =rotse Subtract 5e, and then <C=Sb. Q. E. D.

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Corollary. From hence it is evident, that if two Lines intersect each other, they will make four Angles ; which, being taken together, will always be equal to Four Right-angles.

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THE OREM III. If a Right-line cut (or cross) two parallel Lines, it will make the

opposite Angles equal one to another. (29.6. 1.) Suppose the two Lines A B and H K to be parallel, and the Right-line D G to cut them both at C and n: Upon the Point C (with any Radius) describe a Semicircle ; and with the fame Radius, upon the Point at ni, describe another Semicircle opposite to the first, as in the Figure. Then 'tis plain, and I suppose very easy to conceive, that if the Center G were moy'd along upon the Line DG, until it came to the Center at n, the two Lines A B and H K would meet and concur, viz. become one Line (for parallel Lines are as it were but one broad Line.) Consequently the two Semicircles would also meet, and become one entire Circle, like to that in the laft Demonstration. And therefore the sy s x=ra=re US as before, per And sn= n=5b-sesl last Theoremi.

Q. E. D. Corollary. Hence it follows, that if three, four, or ever so many Parallel lines, are cut or cross'd by one Right-line, all their opposite Angles will be equal.

T HE OR EM IV. The three Angles of every plain Triangle are equal to two Right-angles.

(32. 6. 1.) Consequently, any two Angles of any plain Triangle must needs be less than two Right angles. (17. 6. 1.)

Demons Demonftration. Let the A ABC be propos’d; draw the Right-line H K parallel to the Side A B, juft touching the Vertical Angle C; and upon the fame Angular Point C describe any se SK micircle, and produce the Sides A C, and B C to its Periphery. Then will sb=5B, sa = <A, and

x = C, per laft Theorem. But A sbt sats * = 180°, or two Right-angles : Consequently <B+SA+S C = 180° per Axiom 5. Q. E. D.

Corollary. Hence it follows, that the two acute Angles of every Rightangled Triangle are equal to a Right-angle, or 90°.

Confequently, if one of the acute Angles be given, the other is also given, viz. 90°- the given 5 leaves the other .

THEOREM V. if one Side of any plain Triangle be continued or produced beyond, or !

out of the Triangle, the outward Angle will always be equal to the two inward opposite Angles. (32. e. 1.)

Demonttration. Let the Side A B of the A ABC be produced out of the A, fuppose to D, &c. as in the Fi. gure. Then <z=SA+ *C, for the B+ 5z = 180° per Theorem I, and the

B\

ZD <Bts At5C=180°, per last Theorem. Therefore TB+52=SBSA+SC, per Axiom 5. Subtract TB on both Sides the Æquation, and it will leave sz=5A t5C (per Axiom 2.) Q. E. D.

Confequently, the outward Angle (at 2) of any plain Triangle, mnust needs be greater than either of the inward opposite Angles, viz.. greater than 5 A, or 5C (16. e. 1.)

Corollary. Hence it follows; that if one Angle of any plain Triangle bé given, the Sum of the other two Angles is also given ; for 180° the given the other two 55

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THEOREM VI. In every plain Triangle, equal Sides fubtend (viz. are opposite to j equèl

Angles. (5. é. 1.) Consequently, equal Angles are subtended by equal Sides. (6.6. 1.)

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Corollary. From hence it follows, that the three Angles of an Equilateral Triangle are equal one to another.

THEOREM VII. In every plain Triangle, the longest Side subtends the greatest Angle.

(18.8. 1.) Consequently, the greatest Angle of any plain Triangle is subtended by

the longest Side.

This Theorem is evident by Inspection only: For, let one of the Sides of any plain Triangle (as C B) be produced, suppose to E; join D E with a Right line; then ’ris evident, that beCause G. E is now made longer than the

BL Side B C, therefore the sat D is become larger than it was before by the <BDE: And its plain, the longer the Side CE had been made, the ; at D would have been the more enlarged.

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THEOREM VIII. If the Sides of two Triangles are equal, the Angles opposite to those

equal Sides will be equal. (8.e. 1.) The Truth of this Theorem is evident by the two included Triangles in the 6th Theorem, for they have their respective Sides equal, viz. BC=CD, BA=DA, and C A common to both Triangles. And it is there prov'd, that the opposite to those equal Sides are equal, &c. which needs no further Proof.

Note, The Converse of this Theorem holds not true ; for the Angles of two Triangles may be equal, and their opposite or fubtending Sides unequal; as will appear at Theorem XII.

- Corollary. , Hence it follows, that Triangles mutually equilateral are also mutually equiangular ; and,

That Triangles mutually equilateral are equal one to another. (4. & 26. 6. 1.)

THEOREM IX. An Angle at the Center of any Circle is always double to the Angle at the Periphery, when both the Angles stand upon the same Arch. (20. 6. 3.) This Theorem hath three Varieties or Cafes. .

Demonttration.' Cafe 1. Let the Diameter D A, and the Line D B, be the two Lines which form the 5 D at the Periphery; diaw the Radius B C, then < B C Ä is the sat the Center. But 5 BCA== D for 5B, per Theor. 5. and because DC=BC, therefore <D= B, per Theorem 6. consequently BCA = 2FD.

Cafe 2. Suppose the 5 B C F at the Center to be within the BDF at the Periphery, (as in the annexed Figure,) draw the Diameter DA;, then the BCA=2<BDA2..

" Delo and the FCA=2-FDA Peinago

F . add these two Equations together,

Then

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