fore equal to the radiant; but the radiant is equal to C, wherefore AE is (Ax. 1.) equal to C.

From the same point on the same side of a straight line only one perpendicular to it can be drawn.

Let the straight line CD be perpendicular to AB at the point C, then any other straight line CE, drawn from C on the same side of AB, cannot be perpendicular to it.

D E

For as by hypothesis CD is perpendicular to AB, the angle ACD is equal (Def. 9) to the angle BCD; but BCD is greater (Ax. 8) than the angle BCE, therefore the angle ACD is greater than BCE. Again, the angle ACE is greater than A C B (Ax. 8) ACD; but ACD, as already proved, is greater than BCE, much more then is the angle ACE greater than BCE; wherefore the straight line CE does not make equal adjacent angles with AB, and therefore cannot (Def. 9) be perpendicular to it.

PROPOSITION IV. THEOR.

All right angles are equal to one another.

Let CD be at right angles to AB, and GH at right angles to EF, the four right angles which are thus formed at the points C and G are all equal to one another.

Fig. 1.
D

Fig. 2.

LH K

For the angle ACD is equal to BCD, and the angle EGH to FGH; for if they were not, they could not (Def. 9) be right angles. But the angles at C are also equal to the angles at G. For suppose Fig. 1 to be applied to Fig. 2, so that the point C may be on G, and the straight line CA and GE, the line CB shall coincide (Def.

A

B E

3, Cor. 1) with GF, and the line CD shall fall on GH. For if CD do not fall on GH, it must fall either on the one side of GH or on the other, in the position GK or GL. But it cannot fall in the position GK; for then there would be two perpendiculars, GH and GK, to the line EF, at the point G, on the same side of it, which (Prop. 3) is impossible. For the same reason, CD cannot fall on GL. Since therefore CD can neither fall on the one side of GH nor the other, CD must fall upon it; and therefore the right angles at the points C and G are all equal, viz., the angle ACD to EGH, and BCD to FGH.

PROPOSITION V. THEOR.

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

Let the straight line CD make with AB on the same side of it the angles ACD and BCD, these are either two right angles, or are together equal to right angles.

A

E

B

For if the angle ACD be equal to BCD, each of them is (Def. 9) a right angle, and CD is perpendicular to AB. From the point C draw any other straight line CE on the same side of AB. Because CD is perpendicular to AB at the point. C, the line CE cannot be (Prop. 3) also perpendicular at the same point to AB on the same side of it; and therefore (Def. 9) the angles ACE and BCE are unequal. Now the angle BCD is equal (Ax. 9) to the two angles BCE and ECD; to each of these equals add the angle ACD, and the two angles BCD and ACD are equal (Ax. 2) to the three BCE, ECD, and DCA; but the angles ECD and DCA are equal to ECA, and, by substitution, the angles ACD and BCD are equal to the angles ACE and BCE; but the angles ACD

C

and BCD are two right angles; therefore the angles ACE and BCE are together equal to two right angles.

Cor. 1. The sides of a right angle are mutually perpendicular. For if either side of a right angle be produced through the vertex, the angle which it makes with the other side is a right angle.

Cor. 2. If a straight line be perpendicular to another on one side, it is also perpendicular to it on the other.

Cor. 3. All the simple angles at the same point in a straight line, and on the same side of it, are together equal to two right angles.

PROPOSITION VI. THEOR.

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

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

For if BD be not in the same straight line with CB, the production of CB must lie either above or below it, in the position BE or BF; but it cannot lie above it, in the position BE, because the straight line AB makes angles with Cthe straight line CBE on one side of it; the

A

F

angles ABC and ABE are together equal (Prop. 5) to two right angles; but the angles ABC and ABD are also together equal (Hyp.) to two right angles; therefore the angles ABC and ABE are equal to the angles ABC and ABD; take away the common angle ABC, and there remains (Ax. 3) the angle ABE, equal to ABD, the less to the greater, which is impossible. The production of CB cannot therefore

lie above BD. In the same manner, it may be demonstrated that it cannot lie below BD, in the position BF. Since therefore the production of CB can neither lie above nor below BD, it must lie upon BD; and therefore CB and BD are in the same straight line.

PROPOSITION VII. THEOR.

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

Let the two straight lines AB and CD cut one another in the point E, the angle AEC is equal to DEB, and the angle CEB to AED.

E

B

For the angles CEA and AED, which c the straight line AE makes with CD, are A together equal (Prop. 5) to two right angles; and the angles AED and DEB, which the straight line DE makes with AB, are together equal to two right angles; therefore the two angles CEA and AED are equal to the two AED and DEB: take away the common angle AED, and the remaining angle CEA is equal (Ax. 3) to the remaining angle DEB. In like manner, it may be demonstrated that the angle CEB is equal to AED.

Cor. 1. From this it is evident, that if two straight lines cut one another, the angles which they make at the point of their intersection are together equal to four right angles.

Cor. 2. Hence also all the simple angles made by any number of straight lines meeting in one point are together equal to four right angles.

PROPOSITION VIII. THEOR.

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

Let ABC be a triangle, any two of its sides are together greater than the third.

A

For as a straight line is (Def. 3, Cor. 2) the shortest distance between two points, it is manifest that the crooked line, composed of BA and AC, is greater than the straight line BC, and, for the same reason, AB and BC are greater than AC, and AC and CB greater than AB; and therefore any two sides of a triangle are together greater than the third side.

B

C

PROPOSITION IX. LEMMA.

The circumferences of circles cannot cut one another in more than two points.

Let A and B be two circles, and let their circumferences cut one another in the point C, they cannot cut one another in more than a second point D.

For if the circumference of A can cut that of B in more points than D, let these be E and F; let G be the centre of the circle A, and join GE and GF; let GE meet the circumference of A in the point H, and produce GF to meet it in K.

F H

B

Because G is the centre of the circle A, the line GH, drawn from it to the circumference, is (Def. 34) a radius of A, and the point I is in its circumference. But GE is greater (Ax. 9) than GH; but if the circumference of A passed through the point E, the line GE would also be a radius of A, and equal to GII, which it is not; therefore the circumference of A does not pass through the point E; and GE being greater than the radius GH, the point E is without the circle A. Again, GK is a radius of the circle A, and GK is greater than GF; but if the circle A passed through the point F, the line GF would be equal to GK, which it is not; therefore the cir