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.).

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.

D

A

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

B

C

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.

equal respectively to the two B angles E, F, and interjacent

CE

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 ACB, a part to the whole, which is absurd; therefore, AB cannot be longer than DE, and in like manner it may 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.

be

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.

For if these lines are not parallel, let them meet in some point I, and through M, the middle of BC, draw

E

B M

C

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.

Let the two points F, G in the CF G D straight line CD be unequally distant from the straight line AB, then the line CD by being

Ꮭ

produced, shall approach nearer and nearer to AB or its prolongation till it meets it.

For CDE cannot, at any point of its progress, discontinue to approach, and then proceed at equal distance from AB, as is manifest from the corollary to Prop. V.; still less could it, after approaching, reverse its direction, and recede from the line ABF; for it is the obvious characteristic of a straight line to preserve invariably one direction, however far it be extended, so that if the straight line CD be directed to a point infinitely distant, it will, by being infinitely prolonged, at length arrive at, and terminate in, that point, and must necessarily in its progress onward pass through every straight line crossing its path; and it is plain, that its continual approach toward any straight line is an indication that such line does cross its path, and must therefore be eventually intersected by it.

Cor. 1. It therefore follows that if one straight line be parallel to another, it must be every where equidistant from it. Cor. 2. Hence from the same point more than one parallel to a straight line cannot be drawn.

Cor. 3. And therefore two straight lines that are parallel to the same straight line are parallel to each other, for if they could meet, the last corollary would be contradicted.

PROPOSITION XIV. THEOREM. (Converse of Prop. XII.)

If a straight line intersect two parallel lines it will make the alternate angles equal.

Let AD intersect the parallels EF, GH, in B and C; the alternate angles EBC, HCB, are equal.

For if the angle EBC is not equal to the angle HCB, let the angle eBC be equal thereto, then since the alternate angles eBC, HCB are equal, Be is parallel to GH (Prop. XII.); Abut by hypothesis, BE is also parallel to GH, so that from the same point B, two parallels to GH are drawn, which is impossible (Prop. XIII.

e E

e

G

B

D

H

Cor. 2.). Hence the alternate angles EBC, HCB are equal.

Cor. 1. Therefore a straight line intersecting two parallels makes the exterior angle equal to the interior opposite one, on the same side of the cutting line.

Cor. 2. The alternate exterior angles (ABE, DCH) are also equal.

Cor. 3. The two exterior angles also (ABE, DCG) on the same side are together equal to two right angles.

Cor. 4. Also, a line intersecting two parallels makes the two interior angles on the same side together equal to two right angles.

Cor. 5. A line perpendicular to one of two parallels is perpendicular to the other.

Cor. 6. Therefore, if to each of two parallels perpendiculars be drawn, these perpendiculars shall be parallel; for by last corollary they are all perpendicular to the same line.

Scholium.

These corollaries prove the converse of the corollaries to proposition XII.

PROPOSITION XV. THEOREM.

Two angles are equal if the sides of the one are parallel, each to each, to the sides of the other, and at the same time lie each upon that side of the line joining their vertices which the parallel side lies on, or else each upon the opposite side of that line. Let the sides of the angles BAC,DEF, be parallel each to each, and let AB lie upon the same side of AE as the parallel side ED, and let

B

D

D

D

AC also lie upon the same side of this line as EF, the two angles are equal.

For since BA, DE are parallel, the angles BAE, DEG are equal (Prop. XIV. Cor. 1.). For similar reasons the angles CAE, FEG are also equal; therefore the angle BAC is equal to the angle DEF. If the parallel sides of the two angles lie upon contrary sides of AE, as Ab, ED, Ac, EF, it is obvious the angles will still be equal, for the angle bAC is equal to the angle BAC.