THEOREM I.

If a Right-line ftand 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.)

Demonstration.

Suppofe the Lines to be AB and DC, meeting in the Point at C: Upon C defcribe any Circle at

pleafure; then will the Arch A D

be the Measure of the

b, and the
e;

Arch D B the Measure of

but the Arches A D+D B=180°,

viz. they compleat the Semicircle.

be

C

Confequently the <6+e=180°. Which was to be provid.

Corollaries.

e90°;

1. Hence it follows, that if the b 90° then e 90°; but ifb be obtufe, then thee will be acute, &c.

From hence it will be eafy to conceive, that if feveral Righte lines ftand upon, or meet with any Right-line at one and the fame Point, and on the fame Side, then all the Angles taken toge ther will be 180°, viz. Two Right-angles.

THEOREM II.

If two Angles interfect (i. e. cut or cross) each other, the two oppofite Angles will be equal. (15. e. 1.)

Demonftration.

Let the two Lines be A B and

D E, interfecting each other in the
Center C.

Then be=180°

And 180° per laft.

Confequently

b+<e=<6+

<a, per Axiom 5.

Subtract on both Sides of

the Equation, and it will leave

<e=<a.

Again, be 180°, as before; and <e+<C= 180°, confequently<e+C=<b+<e. Subtract <e, and then <C=<b. Q. E. D.

Corol

Corollary.

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

THEOREM III.

If a Right-line cut (or cross) two parallel Lines, it will make the oppofite Angles equal one to another. (29. c. 1.)

Suppose the two Lines A B and H K to be parallel, and the Right-line DG to cut them both at C and n: Upon the Point C (with any Radius) defcribe a Semicircle; and with the fame Radius, upon the Point at n, describe another Semicircle oppofite to the first, as in the Figure. Then 'tis plain, and I fuppofe very easy to conceive, that if the Center C were mov'd along upon the Line DG, until it

4

B

m/x

H

y/n

K

came to the Center at n, the two Lines AB and HK would meet and concur, viz. become one Line (for parallel Lines are as it were but one broad Line.) Confequently the two Semicircles would also meet, and become one entire Circle, like to that in the laft Demonftration.

And therefore they

And

<a=<e} as before, per m<b<elaft Theorem.

Corollary.

Q. E. D.

Hence it follows, that if three, four, or ever fo many Parallel lines, are cut or crofs'd by one Right-line, all their oppofite Angles will be equal.

THEOREM IV.

The three Angles of every plain Triangle are equal to two Right-angles.

(32. e. 1.)

Confequently, any two Angles of any plain Triangle must needs be lefs than two Right angles. (17. e. 1.)

Demons

Demonftration.

Let the AABC be propos'd; draw the Right-line HK parallel to the Side AB, juft touching the Vertical Angle C; and upon the fame Angular Point C defcribe any Semicircle, and produce the Sides A C, and B C to its Periphery. Then will b=B, <=<A, and x =C, per laft Theorem. But <6+ <a+ x = 180°, or

a

B

two Right-angles: Confequently <B+<^+<C = 180° per Axiom 5. QE. 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 alfo given, viz. 90°- the given 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 oppofite Angles. (32. e. 1.)

Demonstration.

Let the Side AB of the AABC be produced out of the A, fuppofe to D, &c. as in the Figure. Then 2=1+

C, for the <B+<%= 180° per Theorem 1. and the <B+<A+C=180°, per laft Theorem.

Therefore

BZ -D

<B+<%=<B+<A+C, per Axiom 5. Subtrac B on both Sides the Equation, and it will leave 24 +C (per Axiom 2.) Q. E. D.

Confequently, the outward Angle (at z) of any plain Triangle, muft needs be greater than either of the inward oppofite Angles, viz. greater than A, or C (16. e. 1.)

Corollary.

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

T H E 0.

THEOREM VI.

In every plain Triangle, equal Sides fubtend (viz. are oppofite to) equal Angles. (5. e. 1.)

Confequently, equal Angles are fubtended by equal Sides. (6. e. 1.)

Demonstration.

=

Suppofe the ABCD to be an Ifofceles A; that is, let B C C D. Bifect the C, or (which is all one) make C A perpendicular to BD; then will the (viz.BAC and

angles.

Therefore {

Σ

on each Side of it DAC) be Right

C+B=90°

C

C+<D=90° } per Corol, to Theorem 4. Confequently,< C + < B = { <C+<D, per Axiom 5. SubtractC from both Sides of the Equation, and it will leave BD, per Axiom 2. Q. E. d.

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 fubtends the greatest Angle. (18. e. 1.)

Confequently, the greatest Angle of any plain Triangle is fubtended by the longest Side.

B

This Theorem is evident by Infpection only: For, let one of the Sides of any plain Triangle (as CB) be produced, fuppofe to E; join D E with a Right line; then 'tis evident, that becaufe CE is now made longer than the Side B C, therefore the at 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.

THE O

THEOREM VIII.

If the Sides of two Triangles are equal, the Angles opposite to thofe equal Sides will be equal. (8. e. 1.)

The Truth of this Theorem is evident by the two included Triangles in the 6th Thearem, for they have their refpective Sides equal, viz. BC=C D, BADA, and C A common to both Triangles. And it is there prov'd, that the oppofite to those equal Sides are equal, &c. which needs no further Proof.

Note, The Converfe of this Theorem holds not true; for the Angles of two Triangles may be equal, and their oppofite 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. e. 1.)

THEOREM IX.

An Angle at the Center of any Circle is always double to the Angle at the Periphery, when both the Angles ftand upon the fame Arch. (20. e. 3.) This Theorem hath three Varieties or Cafes.