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right angles (Hyp.); therefore the angles CBA, ABE are equal to the angles CBA, ABD: From each of these equals take away the common angle CBA; then the remaining angle ABE is equal to the remaining angle ABD (Ax. 3), the less to the greater--which is absurd 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: Therefore BD is in the same straight line with BC.

Wherefore, If at a point &c.

PROP. XV.

Q. E.D.

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 BED, and the angle AED to the angle BEC. Because the straight line AE makes with CD the angles AEC, AED, these are together

C

equal to two right angles (1.13):

Again, because the straight line DE

A

E

D

makes with AB the angles AED, BED,

these also are together equal to two right angles: But the angles AEC, AED have been shewn to be equal to two right angles; therefore the angles AEC, AED, are equal to the angles AED, BED: From each of these equals take away the common angle AED; then the remaining angle AEC is equal to the remaining angle BED.

And in like manner it may be demonstrated that the angle AED is equal to the angle BEC.

Wherefore, If two straight lines &c. Q. E. D.

COR. 1. From this it is manifest that, if two straight lines cut one another, the angles 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.

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If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

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

Bisect AC in E (1. 10), join BE and produce it to F, making EF equal to BE (1. 3); and join FC.

M

B

E

с

D

Then, because AE is equal to EC, and BE to EF, the two sides, AE, EB, are equal to the two, CE, EF, each to each, and the angle AEB is equal to the angle CEF, because they are opposite vertical angles (1. 15);—therefore the base AB is equal to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; therefore the angle BAE is equal to the angle ECF: But the angle ECD is greater than the angle ECF; therefore the angle ACD is greater than the angle BAE, or BAC.

And, in like manner, if the side BC be bisected and AC be produced to G, it may be demonstrated that the angle BCG, that is, the angle ACD (1. 15.), is greater than the angle ABC.

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Any two angles of a triangle are together less than two right angles.

Let ABC be a triangle: any two of its angles are together less than two right angles.

Produce BC to D: Then, because ACD is the exterior angle of the triangle ABC, ACD is greater than the interior and opposite angle ABC (1. 16): To each of these add the angle ACB; therefore the angles ACD, B ACB, are together greater than the angles ABC, ACB (Ax. 4): But the angles ACD, ACB, are together equal to two right angles (1. 13); therefore the angles ABC, ACB are together less than two right angles. And, in like manner, it may be demonstrated that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles.

Wherefore, Any two angles &c.

Q.E. D.

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The greater side of every triangle is opposite to the greater

angle.

Let ABC be a triangle, of which the side AC is greater than the side AB: the angle ABC is also greater than the angle ACB.

Since AC is greater than AB, cut off AD equal to AB, and join BD: Then, because ADB is the exterior angle of the triangle BDC, it is greater than the interior and opposite angle DCB (1. 16): But the angle ADB

B

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is equal to the angle ABD, because the side AD is equal to the side AB (1. 5); therefore also the angle

ABD is greater than the angle ACB: much more then is the angle ABC greater than the angle ACB. Wherefore, The greater side &c. Q. E.D.

PROP. XIX. THEOR.

The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

Let ABC be a triangle, of which the angle ABC is greater than the angle ACB: the side AC is also greater than the side AB.

B

A

For, if it be not greater, it must either be equal to it, or less: But AC is not equal to AB, because then the angle ACB would be equal to the angle ABC (1. 5); but it is not therefore AC is not equal to AB: Neither is it less, because then the angle ACB would be less than the angle ABC (1. 18); but it is nottherefore AC is not less than AB: And it has been shewn that it is not equal to it; therefore AC is greater than AB.

Wherefore, The greater angle &c. Q.E.D.

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Any two sides of a triangle are together greater than the third side.

Let ABC be a triangle: any two sides of it are together greater than the third side, viz. BA, AC greater than BC, and AB, BC, greater than AC, and AC, CB, greater than AB.

Produce BA to the point D, making
AD equal to AC (1. 3), and join DC.
Then because AD is equal to AC, the
angle ADC is equal to the angle ACD: B

D

But the angle BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle BDC: And because the angle BCD of the triangle BCD is greater than the angle BDC of the same triangle, and that the greater side is opposite to the greater angle (1. 19), therefore the side BD is greater than the side BC: But BD is equal to BA and AC; therefore BA, AC are greater than BC.

And in like manner it may be demonstrated that AB, BC, are greater than AC, and AC, CB, greater than AB.

Wherefore, Any two sides &c.

PROP. XXI.

Q. E.D.

THEOR.

If from the ends of a side of a triangle there 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 the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it: BD and DC are less than the other two sides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC.

A

E

Produce BD to E: Then, because any two sides of a triangle are together greater than the third side (1. 20), the two sides, BA, AE, of the triangle ABE are greater than BE: To each of these add EC; therefore the sides BA, AC are greater than BE, EC: Again, because the two sides, CE, ED, of the triangle CED are greater than CD, to each of these add DB; therefore the sides CE, EB are greater than CD, DB: But it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC.

B

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