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Take away the common angle ABC, and the *3 Ax. remaining angle ABE is equal* to the remaining

angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with BC. And in like manner, it may be demonstrated that no other can be in the same straight line with it but BD, which therefore is in the same straight line with CB. Wherefore, if at a point, &c. Q. E. D.

PROP. XV. THEOR.

If two straight lines cut one another, the vertical,

or opposite angles, shall be equal. Let the two straight lines AB, CD, cut one another in the point E: the angle AEC shall be equal to the angle DEB, and CEB to AED.

Because the straight line AE makes with CD

the angles CEA, AÈD, these 13. 1. angles are together equal * to two right angles. Again, be

A

B cause the straight line DE

makes with AB the angles * 13. 1. AED, DEB, these also are together equal * to

two right angles ; and CEA, AED, have been

demonstrated to be equal to two right angles; +1 Ax. wherefore the angles CEA, AED, are equals to

the angles AED, DEB. Take away the common

angle AED, and the remaining angle CEA is *3 Ax. equal * to the remaining angle DEB. In the

same manner, it can be demonstrated, that the angles CEB, AED are equal. Therefore, if two straight lines, &c. Q. E. D.

Cor. I. From this it is manifest, that, if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles.

Cor. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles.

BOOK I.

PROP. XVII.

21

* 10.1.

PROP. XVI. THEOR.
If one side of a triangle be produced, the exterior

angle is greater than either of the interior op-
posite angles..

Let ABC be a triangle, and let its side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC.

Bisect * AC in E, join А BE and produce it to F,

F and make EF equal t to

13.1. BE, and join FC. Because AE is equal t to

Const. EC, and BE + to EF; AE, B EB are equal to CE, EF, each to each; and the angle

G AEB is equal* to the angle CEF, because they are opposite vertica. angies ; therefore the base AB is equal* to the base CF, * 4. 1. and the triangle AER iy the triangle,CEF, aná the remaining, angles to the renairing angles, each to each, to which the equal sides are opposite: wherefore the angle:BAE is equal 1o the angle ECF: but the angle SCD is greater+then fo°4x. the angle ECF, therefore the anglé ÀCD is greater than BAE. In the sane manzer if ilie) side BC be bisected, and AC be produced to‘G, it may be demonstrated tlial tlie angle BCG, that is, the angle * ACD, is greater than the *15. 1. angle ABC. Therefore, if one side, &c. Q. E. D.

* 15.1.

PROP. XVII. THEOR.
Any two angles of a triangle are together less

than two right angles.
Let ABC be any triangle; any two of its
angles together shall be less than two right
angles.

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Produce BC to D; and be

А cause ACD is the exterior

angle of the triangle ABC, * 16 1. AČD is greater * than the in

terior and opposite angleABC;
to each of these add the angle B

ACB; therefore the angles + 4 Ax. ACD, ACB are greater | than the angles ABC, *13. 1. ACB: but ACD, ACB are together equal * to

two right angles; therefore the angles ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as also CÁB, ABC, are less than two right angles. Therefore any two angles, &c. Q. E. D.

angle BĆA

PROP. XVIII. THEOR.
The greater side of every triangle is opposite lo

the greater angle.
Let ABC be a triangle, of

A
which the side AC is greater
than the side AB:'{lie'aggle
ABC shu bo greater thavi the

B
Becauze AC is gicater than
AB, make * AD'equal to AB, and join BD: and

because ADB is the exterior angle of the triangle • 16. I. BDC, it is greater* than the interior and oppo

şita anglo DCB; bụt ADB is equal * to ABD, + Const. because the siáe AB is equal to the side AD;

therefore the angle ABD is likewise greater than the angle ACB: therefore much more is the angle ABC greater than ACB. Therefore the greater side, &c. Q. E. D.

*3.1.

*5. 1.

*

PROP. XIX. THEOR.
The greater angle of every triangle is subtended

by the greuter side, or has the greater side
opposite to it.
Let ABC be a triangle of which the angle

. 18. 1.

ABC is greater than the angle BCA: the side
AC shall be greater than the side AB.

For, if it be not greater, AC must either be equal to AB, or less than it: it is not equal, because then the angle ABC would be equal * * 5. 1 to the angle CAB ; but it is f not; therefore + Hyra AC is not equal to AB: neither is it less; because then the angle ABC would be less* than the angle ACB: but it is not; therefore the side AC is

B
not less than AB ; and it has
been shown that it is not equal to AB; there-
fore AC is greater than AB. Wherefore the
greater angle, &c. Q. E. D.

PROP. XX. THEOR.
Any two sides of a triangle are together greater

than the third side.
Let ABC be a triangle : any two sides of it
together shall be greater than the third side,
viz. the sides BA, AC greater than the side
BC; and AB, BC greater than AC; and BC,
CA greater than AB.
Produce BA to the point

D
D, and make * AD equal to
AC; and join DC.

Because DA is equal to AC, the angle ADC is equal* B to ACD; but the angle BCD is greatert than the angle ACD: there- 19 Ax. fore the angle BCD is greater than the angle ADC: and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater* angle is subtended by the *19. 1. greater side; therefore the side DB is greater than the side BC: but DB is equal I to BA and AC; therefore the sides BA, AC are greater than BC. In the same manner it may be

Because AD is equal t to AC, add BA to each, therefore + Conse the whole BD is equal t to the two BA and AC.

12 Ax.

* 3.1.

5.1.

D

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demonstrated, that the sides AB, BC are
greater than CA, and BC, CA greater than AB.
Therefore any two sides, &c. Q. E. D.

PROP. XXI. THEOR.
If from the ends of the side of a triangle, there

Le 1,8.
be drawn two straight lines to a point within
the triangle, these shall be less than the other
two sides of the triangle, but shall contain a
greater angle.

Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle : BD, and DC shall be less than the other two sides BA, AC of the triangle, but shall contain an angle BDC

byta to B. greater than the angle BAC.

Produce BD to E; and because two sides of a * 20. 1. triangle* are greater than the third side, the two

sides BA, AĚ, of the triangle ABE are greater

than BE. To each of these add EC; therefore 44. Ax. the sides BA, AC are greatert

than BE, EC. Again, because E

the two sides CE, ED of the † 20. 1. triangle CED are greatert than CD, add DB to each of

therefore the sides CE, B † 4 Ax. EB, are greatert than CD,

DB : but it has been shown that BA, AC are
greater than BE, EC; much more then are
BA, AC, greater than BD, DC.

Again, because the exterior angle of a triangle*
is greater than the interior and opposite angle,
the exterior angle DC of the triangle CDE is
greater than CED; for the same reason, the ex-
terior angle CEB of the triangle ABE is greater
than BAC: and it has been demonstrated that
the angle BDC is greater than the angle CEB;
much more then is the angle BDC greater than
the angle BAC. Therefore, if from the ends of
&c. Q. E. D.

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