2. And that it may be either increased till it be equal to a greater straight line, or diminished till it be equal to a less. 3. Grant also, that an angle may be increased till it be equal to a greater angle, or diminished till it be equal to a less.

4. And lastly, that from a point either within or without a straight line, a perpendicular thereto may be drawn.

It is necessary to remark, that in the first eight books of these elements, the lines concerned in each proposition are supposed to be all situated in the same plane.

PROPOSITION I. THEOREM.

From the same point, in a given straight line, more than one perpendicular thereto cannot be drawn.

Let BD be perpendicular to the straight line AB, or AC, BC being the production of AB, and if the truth of the theorem be denied, let some other line, as BE drawn from the same point B, be also perpendicular to AC.

Then because the angles ABD, CBD are equal, (Def. 3.) the angle ABD must be greater than the angle EBC (Ax. 6.). But BE is perpendicular to AC, by hypothesis, therefore the angle EBC must be equal to the angle ABE

B

E

(Def. 3.). It follows, therefore, that the angle ABD is greater than the angle ABE, a manifest absurdity; therefore BE cannot be perpendicular to AC.

PROPOSITION II. THEOREM.

All right angles are equal to each other.

Let ABC be a right angle, and DEF any other right angle, then, if it be denied that these two angles are equal to each other, one of them, as ABC, must be supposed greater than the other, so that DEF must be equal to some portion of ABC. Let ABf represent that portion, then, because ABƒ is a right angle, Bf is perpendicular to AB (Def. 4.); but ABC is also a right angle, therefore BC is likewise perpendicular to AB, that is from the A same point B in the straight line AB, two perpendiculars thereto are drawn, which is impossible. (Prop. I.) Therefore ABC cannot be greater than DEF, and in a similar manner it may be proved that DEF cannot be greater than ABC; the two angles are, therefore, equal.

B D

E

PROPOSITION III. THEOREM.

The adjacent angles which one straight line makes with another which it meets, are together equal to two right angles.

Let the straight line DB meet AC in B, making adjacent angles ABD, CBD; these angles shall together be equal to two right angles.

For, let BE be perpendicular to ABC, then the angle ABD is equal to the right angle ABE, together with the angle EBD, and this angle EBD, together with DBC, make up the other right angle EBC; consequently the sum of the angles ABD, CBD, is equal to two right angles.

A

Corollary 1. Hence, if either side of a right angle be produced through the vertex, the adjacent angle formed will be right.

Cor. 2. Therefore the sides of a right angle are mutually perpendicular.

E

Cor. 3. The sum of all the angles formed by straight lines drawn on the same side of another straight line from any point in it, is equal to two right angles; for, be these angles ever so numerous, they are evidently only subdivisions of the two right angles, which a perpendicular from the point forms with the adjacent portions of the line.

A

C

PROPOSITION IV. THEOREM. (Converse of Prop. III.) If, to the point where two straight lines meet, a third be drawn, making with them adjacent angles, which are together equal to two right angles; the two lines so meeting form but one continued straight line.

Let the two straight lines AB, CB, meet in the point B, to which let a third DB be drawn, so that the adjacent angles DBA, DBC, may together be equal to two right angles, then will ABC be one straight line.

For, if it be denied, let BF, and not BC, be the continuation of AB; then the angles ABD, FBD, are together equal to two right angles (Prop. III.). But the angles AABD, CBD, are together also equal to two

right angles by hypothesis; hence the angle FBD is equal to the angle CBD, a part to the whole, which is impossible; therefore BC is the production of AB.

PROPOSITION V. THEOREM.

Two straight lines having two points common to both, form but one continued straight line.

Let A, B, be the two points through which two straight lines pass, then they must necessarily coincide between A and B (Def. 1.); but if they do not coincide throughout, let ACD be the direction of one, and ACE that of the other; and at the point C, where they separate, let there be CF perpendicular to ACD.

B

Then, because CF is perpendicular to the straight line ACD, FCD, is a right angle (Def. 3.); and since by hypothesis ACE is a straight line, and FCA a right angle, the angle FCE is also a right angle (Prop. III. Cor. 1.), but all right angles are equal to each other (Prop. II.); therefore the whole angle FCD is equal to the part FCE, which is absurd.

Cor. Hence it follows that if two points in a straight line be equally distant from another straight line, the former shall be equally distant from the latter throughout; for if an equidistant straight line be drawn through these points, this line and the former will have two points common, they must, therefore, coincide.

PROPOSITION VI. THEOREM.

If two straight lines intersect each other, the opposite angles formed at their intersection will be equal.

If the two straight lines AB, CD, intersect C at E, the opposite angles CEB, AED, will be equal.

Α

E

For the sum of the angles CEA, CEB, is equal to two right angles (Prop. III.). Also the sum of the angles CEA, AED, is equal to two right angles; that is, the sum of the angles CEA, CEB, is equal to the sum of the angles CEA, AED; and taking away from each of these equal sums the common angle CEA, the remaining angles CEB, AED, must be equal.

I a similar manner it might have been shown that the opposite angles CEA, DEB, are equal.

Cor. 1. Hence (Prop. III.) the sum of the angles formed by

the intersection of two straight lines is equal to four right angles.

Cor. 2. And, therefore, the amount of all the angles formed by the meeting of any number of straight lines in the same point is equal to four right angles, since they are only so many subdivisions of the angles formed by the intersection of AB, CD.

PROPOSITION VII. THEOREM. (Converse of Prop. VI.) If the opposite angles formed by four straight lines meeting in a point are equal, these lines shall form but two straight lines.

Let the four straight lines AE, BE, CE, DE, meet in the point D (see the diagram to last proposition), so that the opposite angles CEB, AED, may be equal; and also the other opposite angles CEA, DEB; then AEB and CED shall be straight lines.

For, since the sum of the angles CEA, CEB, is by hypothesis equal to the sum of the angles DEB, DEA; and the sum of all four is, by the corollary to last proposition, equal to four right angles; it follows that each of the above sums must be equal to two right angles, so that the straight line CE makes, with the two AE, BE, adjacent angles, which are together equal to two right angles; therefore AEB is a straight line (Prop. IV.). In a similar manner it may evidently be proved that CED is a straight line; hence the four lines form but two distinct straight lines.

If two sides, and the included angle in one triangle be equal to two sides, and the included angle in another triangle, those triangles shall be equal.

Let the triangles ABC, DEF,

have the sides AB, AC, and the
included angle A in the one equal
to the two sides DE, DF, and B
the included angle Din the other;

then shall the angle B be equal to the angle E; the angle C equal to the angle F, and the side BC equal to the side EF.

For, since AB is equal to DE, and the angle A equal to the angle D, a triangle equal to DEF may be conceived to be