Proof.-1. Because DC is equal to CE (const.), and FC common to the two triangles DCF,

ECF;

2. The two sides DC, CF, are equal to the two sides EC, CF, each to each;

3. And the base DF is equal to the base EF; (const.)

4. Therefore the angle DCF is equal to the angle ECF; (I. 8.)

And they are adjacent angles.

5. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; (def. 10.)

6. Therefore each of the angles DCF, ECF, is a right angle. Conclusion. Therefore from the given point C in the given straight line AB, a straight line FC has been drawn at right angles to AB. Q. E. F.

Corollary. By help of this problem, it may be demonstrated that

Two straight lines cannot have a common segment.

Hypothesis. If it be possible, let the two straight lines ABC, ABD, have the segment AB

common to both of them.

Construction.-From the point B, draw BE at right angles to AB. (I. 11.)

Demonstration. 1. Because ABC is a straight line, the angle CBE is equal to the angle EBA. A

(def. 10.)

E

D

B

2. Also, because ABD is a straight line, the angle DBE is equal to the angle EBA. (def. 10.)

3. Therefore the angle DBE is equal to the angle CBE. The less to the greater; which is impossible.

Conclusion. Therefore two straight lines cannot have a common segment.

PROPOSITION 12.-PROBLEM.

To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without it.

(References-Prop. I. 8, 10; post. 3; def. 1û, 15.) Given.-Let AB be the given straight line, which may be

produced to any length both ways, and let C be a point without it.

Sought. It is required to draw from the point C, a straight line perpendicular to AB.

Construction.-1. Take any point

D upon the other side of AB.

2. From the centre C, at the distance CD, describe the circle EGF, meeting AB in F and G. (post. 3.) 3. Bisect FG in H. (I. 10.)

4. Join CF, CH, CG.

H

E

A F

G B

D

Then CH shall be perpendicular to AB.

Proof.-1. Because FH is equal to HG (const.), and HC common to the two triangles FHC, GHC;

2. The two sides FH, HC, are equal to the two sides GH, HC, each to each;

3. And the base CF is equal to the base CG; (def. 15.) 4. Therefore the angle CHF is equal to the angle CHG (I. 8), and they are adjacent angles.

5. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angies is called a right angle, and the straight line which stands on the other is called a perpendicular to it. (def. 10.)

Conclusion. Therefore, from the given point C, a perpendicular has been drawn to the given straight line AB. Q. E. F.

PROPOSITION 13.-THEOREM.

The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles.

(References-Prop. I. 11; ax. 1, 2; def. 10.)

Hypothesis.—Let the straight line AB make with CD,` upon one side of it, the angles CBA, ABD.

Sequence. These angles shall either be two right angles, or shall together be equal to two right angles.

Demonstration.-1. If the angle CBA be equal to the angle ABD, each of them is a right angle. (def. 10.)

2. But if the angle CBA be not equal to the angle ABD, from the point B, draw BE at right angles to CD. (I. 11.) 3. Therefore the angles CBE, EBD, are two right angles. (def. 10.)

4. Now the angle CBE is equal to the two angles CBA, ABE; to each of these equals add the angle EBD.

5. Therefore the angles CBE, EBD, are equal to the three angles CBA, ABE, EBD. (ax. 2.)

6. Again, the angle DBA is equal to the two angles DBE, EBA; to each of these equals add the angle ABC.

7. Therefore the angles DBA, ABC, are equal to the three angles DBE, EBA, ABC. (ax. 2.)

8. But the angles CBE, EBD, have been shown to be equal to the same three angles; (dem. 5.)

And things which are equal to the same thing, are equal to one another;

9. Therefore the angles CBE, EBD, are equal to the angles DBA, ABC. (ax. 1.)

10. But the angles CBE, EBD, are two right angles. (dem. 3.) 11. Therefore the angles DBA, ABC, are together equal to two right angles. (ax. 1.)

Conclusion. Therefore, the angles which one straight line, &c. Q. E. D.

PROPOSITION 14.-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.

(References-Prop. I. 13; ax. 1, 3.)

Hypothesis. 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, together equal to two right angles.

Sequence.-BD shall be in the

same straight line with BC.

(False Hypothesis.)-For if BD be

not in the same straight line with

BC, let BE be in the same straight ċ

line with it.

B

E

D

Demonstration.-1. Because CBE is assumed to be a

straight line, and AB meets it in B,

2. Therefore the adjacent angles ABC, ABE, are together equal to two right angles. (I. 13.)

3. But the angles ABC, ABD, are also together equal to two right angles. (hyp.)

4. Therefore the angles ABC, ABE, are equal to the angles ABC, ABD. (ax. 1.)

5. Take away the common angle ABC,

6. The remaining angle ABE is equal to the remaining angle ABD (ax. 3), the less to the greater; which is impossible. 7. Therefore BE is not in the same straight line with BC. 8. And, in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD.

9. Therefore BD is in the same straight line with BC. Conclusion. Therefore, if at a point, &c. Q. E. D.

PROPOSITION 15.-THEOREM.

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

(References-Prop. I. 13; ax. 1, 3.)

Hypothesis.-Let the two straight lines AB, CD, cut one

another in the point E.

Sequence. The angle AEC shall be equal to the angle DEB, and the angle CEB to the angle AED.

Demonstration.-1. Because the straight line AE makes with CD, the angles CEA, AED, these

angles are together equal to two right angles. (I. 13.)

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

3. But the angles CEA, AED, have been shown to be together equal to two right angles.

4. Therefore the angles CEA, AED, are equal to the angles AED, DEB. (ax. 1.)

5. Take away the common angle AED,

6. The remaining CEA is equal to the remaining angle DEB. (ax. 3.)

7. In the same manner it can be shown that the angles CEB, AED, are equal.

Conclusion. Therefore, if two straight lines, &c. Q. E. D.

B

Corollary.-1. 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.

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

PROPOSITION 16.-THEOREM.

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

(References-Prop. I. 3, 4, 10, 15; ax. 9.)

Hypothesis. Let ABC be a triangle, and let its side BC be produced to D.

Sequence. The exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC.

Construction.-1. Bisect AC in E. (I. 10.)

2. Join BE, and produce it to F, making EF equal to BE (I. 3), and B join FC.

Demonstration.-1. Because AE

is equal to EC, and BE equal to EF,

(const.)

A

2. AE, EB, are equal to CE, EF, each to each.
3. And the angle AEB is equal to the angle CEF,

they are opposite vertical angles. (I. 15.)

D

because

4. Therefore the base AB is equal to the base CF. (I. 4.) 5. And the triangle AEB to the triangle CEF. (I. 4.) 6. 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. (I. 4.) 7. But the angle ECD is greater than the angle ECF. (ax.9.) 8. Therefore the angle ACD is greater than the angle BAE. 9. In the same manner if BC be bisected, and the side AC he produced to G, it may be proved that the angle BCG (or its equal ACD), is greater than the angle ABC.

Conclusion.-Therefore, if one side, &c. Q. E. D.