Sidebilder
PDF
ePub

CBA, ABD; these are either two right angles, or are together equal to two right angles.

E

B

[ocr errors]

B

D

D

C For, if the angle CBA be equal to ABD, each of them is a right angle (Def. 10.).

But if not, from the point B draw BE at right angles (I. 11.) to CD; therefore (Def. 10.)

1. The angles CBE, EBD, are two right angles; and because CBE is equal to the two angles CBA, ABE, together, add the angle EBD to each of these equals; therefore (Ax. 2.)

2. The angles CBE, EBD, are equal to the three angles CBA, ABE, EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC; therefore (Ăx. 2.)

3. The angles DBA, ABC, are equal to the three angles DBE, EBA, ABC; but the angles CBE, EBD, have been demonstrated to be equal to the same three angles; and things that are equal to the same are equal (Ax. 1.) to one another; therefore

4. The angles CBE, EBD, are equal to the angles DBA, ABC; but CBE, EBD, are two right angles; therefore (Ax. 1.)

5. DBA, ABC, are together equal to two right angles. Wherefore, the angles which one straight line, &c. Q.E.D.

PROP. XIV.-THEOREM.

If at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

At the point B, in the straight line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD, equal together to two right angles. BD is in the same straight line with CB.

B

D For, if BD be not in the same straight line with CB, let BE be in the same straight line with it; therefore, because the straight line AB makes angles with the straight line CBE, upon one side of it, (I. 13.)

1. The angles ABC, ABE, are together equal to two right angles; but the angles ABC, ABD are likewise together equal to two right angles (Hyp.); therefore (Ax. 1.)

2. The angles CBA, ABE, are equal to the angles CBA, ABD. Take away the common angle ABC, and (Ax. 3.)

3. The remaining angle ABE is equal to the remaining angle ABD, the less to the greater, which is impossible; therefore

4. 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

5. BD is in the same straight line with CB. Wherefore, if at a point, &c. Q.E.D.

PROP. XV.-THEOREM.

If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

Let the two straight lines 4B, CD, cut one another in the point E; the angle AEC shall be equal to the angle DEB, and CEB to AÈD.

B

A А

Because the straight line AE makes with CD the angles CEA, AED, (I. 13.)

1. CEA, AED, are together equal to two right angles. Again, because the straight line DE -makes with AB the angles AED, DEB, (I. 13.)

2. AED, DEB, are together equal to two right angles ; and CEA, AED, have been demonstrated to be equal to two right angles; wherefore (Ax. 1.)

3. The angles CEA, AED, are equal to the angles AED, DEB.

Take away the common angle AED, and (Ax. 3.)

4. The remaining angle CEA is equal to the remaining angle DEB. In the same manner it can be demonstrated that

5. The angles CEB, AED, are equal. Therefore, 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.

PROP. XVI.-THEOREM. 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, BAČ.

A

[ocr errors]

D

B

Bisect (I. 10.) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G. Because is equal to EC, and BE to EF;

1. AE, EB, are equal to CE, EF, each to each; and (I. 15.)

2. The angie AEB is equal to the angle CEF, because they are opposite vertical angles; therefore (I. 4.)

3. 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: wherefore,

4. The angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore

5. The angle ACD is greater than BAE. In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG that is (I. 15.)

6. The angle ACD is greater than the angle ABC. Therefore if one side, &c. Q.E.D.

PROP. XVII.-THEOREM. Any two angles of a triangle are together less than two right angles.

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

А

C

B

D Produce BC to D; and because ACD is the exterior angle of the triangle ABC, (I. 16.)

1. ACD is greater than the interior and opposite angle ABC; to each of these add the angle ACB; therefore

2. The angles ACD, ACB, are greater than the angles ABC, ACB; but (I. 13.)

3. ACD, ACB, are together equal to two right angles; therefore

4. The angles ABC, BCA, are less than two right angles. In like manner, it may be demonstrated, that

5. BAC, ACB, as also CAB, ABC, are less than two right angles. Therefore any two angles, &c. Q.E.D.

ر

PROP. XVIII.-THEOREM.

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 BCA.

D

B

Because AC is greater than AB, make (I. 3.) AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, (I. 16.)

1. ADB is greater than the interior and opposite angle DCB;

but (I. 5.)

2. ADB is equal to ABD, because the side AB is equal to the side AD; therefore likewise

3. The angle ABD is greater than the angle ACB, wherefore much more

4. The angle ABC is greater than ACB. Therefore the greater side, &c. Q.E.D.

PROP. XIX.-THEOREM. 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 BCA; the side AC is likewise greater than the side AB.

B 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 (I. 5.) to the angle ACB; but it is not; therefore

1. AC is not equal to AB; neither is it less; because then the angle ABC would be less (I. 18.) than the angle ACB ; but it is not; therefore

2. The side AC is not less than AB; and it has been shown that it is not equal to AB; therefore

3. AC is greater than AB. Wherefore the greater angle, &c. Q.E.D.

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

Let ABC be a triangle ; any two sides of it together are 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.

D

B

« ForrigeFortsett »