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.

F

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 at E, the opposite angles CEB, AED, will be equal.

Α'

B

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.

Is 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.

B

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;

A

CE

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

formed, of which AB shall be one side, and A an angle; now it is obvious that the side of this triangle, which corresponds to DF, must fall upon AC, otherwise the angles A and D would be unequal; nor can this side extend beyond or fall short of the point C, for DF is equal to AC. The two extremities, therefore, of the base would coincide with the points B, C; the two bases, therefore, would coincide throughout (Prop. V.), so that the triangle so formed would entirely coincide with the triangle ABC, and it is at the same time equal to the triangle DEF; hence the triangles ABC, DEF, are equal.

Cor. If a perpendicular from one of the angles of a triangle to the opposite side bisect that side; it shall also bisect the angle, and the sides containing that angle shall be equal, that is, the triangle will be either isosceles or equilateral.

PROPOSITION IX. THEOREM.

The angles opposite the equal sides of an isosceles triangle are equal.

Let the sides AB, AC, of the triangle ABC be equal, then will the angle C be equal to the angle B.

For, let AD be the line bisecting the angle A; then, in the two triangles ABD, ACD, two sides AB, AD, and the included angle in the one are equal to the two sides AC, AD, and the included angle in the other; hence the angle B is equal to the angle C (Prop. VIII.).

B

D

Cor. 1. It also follows (Prop. VIII.) that BD is equal to CD, and that the angle ADB is equal to the angle ADC; therefore the line bisecting the vertical angle of an isosceles triangle bisects the base at right angles; and conversely, the line bisecting the base of an isosceles triangle at right angles, bisects also the vertical angle.

Cor. 2. Every equilateral triangle is also equiangular.

PROPOSITION X. THEOREM.

(Converse of Prop. IX.)

If two angles of a triangle are equal, the opposite sides

are equal.

In the triangle ABC let the angles ABC, ACB, be equal; then, if it be supposed that one of the opposite sides as AB is longer than the other AC, let BD be equal to AČ; then the triangle DBC is obviously less than the triangle B

A

ABC. But, since CB, BD, and the included angle, are equal

to BC, CA, and the included angle, by hypothesis, it follows that the same triangles are equal (Prop. VII.), which is impossible; therefore AB cannot be longer than AC, and in a similar manner it may be shown that AC cannot be longer than AB; therefore these two sides are equal.

Cor. Therefore every equiangular triangle is equilateral.

Two triangles are equal, if two angles and the interjacent side in the one, are equal to two angles and the interjacent side in the other.

Let the triangles ABC, DEF, have the two angles ABC, ACB, and the inter

jacent side BC, in the one, equal respectively to the two angles E, F, and interjacent

side EF in the other, the two triangles will be equal.

It will be necessary only to show that the side AB must be equal to the side DE (Prop. VII.). If this equality be denied, let one of these sides as AB be supposed longer than the other, and let BG be equal to ED. Join GC, then, since GB, BC, and the included angle B, are respectively equal to DE, EF, and the included angle E, the angle BCG must be equal to the angle F (Prop. VII.); but by hypothesis, the angle F is equal to the angle ACB; hence, then the angle GCB is equal to the angle AB, a part to the whole, which is absurd; therefore, AB cannot be longer than DE, and in like manner it may be shown that DE cannot be longer than AB; AB is therefore equal to DE, and consequently the triangle ABC is equal to the triangle DEF.

.

Cor. From this proposition immediately follows the converse of the corollary to proposition VIII., viz if a perpendicular from one of the angles of a triangle to the opposite side bisect this angle, it shall also bisect the side on which it falls, so that the sides including the proposed angle must be equal. (Prop. VIII. Cor.)

PROPOSITION XII. THEOREM.

If a straight line intersect two others, and make the alternate angles equal, the two lines shall be parallel. Let the straight line AD intersect the two straight lines EF, GH, making the alternate angles EBC, HCB equal, then is EF parallel to GH.

B M

G

For if these lines are not parallel, kustulong ontopische d let them meet in some point I, and anal through M, the middle of BC, draw IK, making MK equal to IM, and join CK. Then the triangles IMB, KMC have the two sides IM, MB, and the included angle in the one equal respectively to the two sides KM, MC, and the included angle in the other; hence the angle IBM is equal to the angle KCM (Prop. VIII.); but by hypothesis, the angle IBM is equal to the angle HCM; therefore the angle KCM is equal to the angle HCM, so that CK, CH must coincide, that is, the line GH when produced, meets IK in two points I, K, and yet does not coincide with it, which is impossible (Prop. V.): therefore the lines EF, GH, cannot meet, they are consequently parallel.

K

H

Cor. 1. Hence also, if the angles EBC, GCD be equal, the lines EF, GH will be parallel, that is, if a straight line intersecting two others make an exterior angle equal to the interior, opposite one on the same side of the cutting line, the two lines shall be parallel.

Cor. 2. It follows, too, that if the angles ABE, DCH be equal, the lines EF, GH will be parallel, that is, if the alternate exterior angles be equal, the two lines will be parallel.

Cor. 3. Hence, likewise, if the two exterior angles on the same side (as ABE, DCG) be together equal to two right angles, the two lines will be parallel.

Cor. 4. Also, if the two interior angles on the same side (as EBC, GCB) be together equal to two right angles, the two lines will be parallel.

Cor. 5. Therefore two straight lines perpendicular to a third are parallel.

Scholium.

This last corollary shows the possibility of the existence of parallel lines (Post. 4.), and therefore also of the rhomboid.

PROPOSITION XIII. LEMMA.

If two points in a straight line be unequally distant from another straight line, the former, by being produced on the side of the least distance, shall continue to approach nearer and nearer to the latter, or its production, till at length it shall meet it.