AA

Dem. If, in the triangles ABC, DEF, the sides AB, AC be equal to the sides DE, DF, each to each,-but the base BC greater than the base EF; then the angle A is greater than the angle D. For if A be not greater than D, it must be equal to it, or less than it. Equal it is not, because BC is not equal to EF (a): neither is it less, because BC is not less than EF (a). Wherefore, if two sides, &c. Recite (a), p. 24.

Q. E. D.

26 Th. Two triangles are equal to one another, in all their parts, which have two angles and a side in the one equal to two angles and a side in the other; whether the equal sides be adjacent to both, or only to one of the equal angles.

Let ABC, DEF be two triangles, having equal angles at B, E and at C, F.

C

1. Let BC EF, which are adjacent to both the equal angles; then, the side AB, for example, will equal the side DE: but if not equal, let it be greater; so that a part of it, as BG, shall equal ED (a) join GC;-Hence the triangles BCG, EFD are equal, having the angles B, E equal, and the sides BC, BG in the one equal to the sides EF, ED in the other (b). Therefore, the angles BCG and F are equal; but the angles BCA and F are equal, by hypoth. therefore BCG a part, equals BCA the whole, which cannot be admitted. Therefore, AB is not greater than DE.

C E

F B

H

2. Let AB=DE, which are adjacent to the equals B, E, and opposite to C, F: then, the sides BC, EF, for example, are equal; but if not equal, let BC be the greater; so that a part of it, as BH, shall equal EF (a). Hence the triangles ABH, DEF are equal, having he angles B, E equal by hypoth. and the sides BA, BH equal to he sides ED, EF (b): therefore the angle BHA equals the angle F, or its equal C; that is, the exterior equals the interior on the same side (c), which is impossible. Therefore BC is not greater than EF. Wherefore, two triangles are equal in all their parts, &c.

Recite (a), p. 3; (b), p. 4; (c), p. 16.

Q. E. D.

27 Th. If a straight line (EF) fall upon two other straight lines (AB, CD), making the alternate angles (AEF, EFD), equal to each other; these two straight lines shall be parallel.

Argument. For, if AB be not parallel to CD, produce them, and they shall diverge in one direction and meet in the other. Let them meet in the point G: therefore GEF is a triangle, whose exterior angle AEF exceeds its interior opposite angle EFG (a), which were said to be equal. Therefore AB and CD, neither meet nor diverge, by production, as was supposed; but are parallel, (b). Recite (a), p. 16; (b), Note def. 4.

Q. E. D.

28 Th. If a straight line (EF), falling upon other two straight lines (AB, CD), make equal the exterior angle (EGB) to the interior opposite angle (GHD on the same side of the line; or make the interior angles (BGH, GHD) equal to two right angles, the two lines shall be parallel.

Argument 1. Since, by hyp. EGB is equal to GHD, and also to AGH (a); therefore AGH is equal to GHD (b); and they are alternate angles (c); therefore AB is parallel

to CD.

A

C

E

B

2. Since by hyp. BGH and GHD are equal to two right angles, and that AGH and BGH are also equal to two right angles (d), the former two are equal to the latter two (6); then taking out the common angle BGH, there remain AGH equal to GHD (e); and they are alternate angles (c): therefore AB is parallel to CD.

Wherefore, if a straight line, &c.

Q. E. D.

Recite (a), p. 15; (b), ax. 1; (e), p. 27; (d), p. 13; (e), ax. 3.

29 Th. If a straight line (EF) cut two parallel straight lines (AB, CD), it makes equal 1, the alternate angles; 2, the interior and exterior angles on the same side; and 3, the two interior angles on the same side to two right angles.

1. The alternate angles AGH, GHD are equal, or not; if not, let AGH be the greater, and add BGH to each; then AGH and BGH exceed BGH and GHD (a): but AGH and BGH are equal to two right angles (b); therefore BGH and GHD are less than two right angles, and the lines AB, CD will meet towards B, D (c); but they are parallel and cannot meet; therefore the alternate angles AGH, GHD are equal, as stated.

2. The exterior and interior angles EGB, GHD, on the same side of EF are equal: for, as vertical angles EGB, AGH are equal (d); and AGH proves equal to GHD: therefore EGB equals GHD, as stated.

3. The interior angles BGH, GHD are equal to two right angles: for AGH and BGH are equal to two right angles (6); and AGH equals GHD: therefore BGH and GHD are equal to the same, as stated. Wherefore, if a straight line, &c. Recite (a) ax. 4;

(b) p. 13;

(c) Note def. 4;

(d) p. 15.

E

Q. E. D.

30 Th. Straight lines (AB, CD) which are parallel to the same straight line (EF), are parallel to each other.

Let GHK cut the parallels AB, EF, also CD, EF; then AB is par

allel to CD.

For since the parallels AB, EF are cut by

a straight line, the alternate angles AGH, A GHF are equal (a).

Also, because the parallels CD, EF are

E.

cut by a straight line, the exterior angle CGHF equals the interior, opposite angle GKD (b).

B

F

Therefore, each of the angles AGH, GKD, being equal to GHF, are equal to each other (c); and being alternate angles, AB is parallel to CD (a).

Wherefore, straight lines, &c.

Recite (a) p. 27; (b) p. 28; (c) ax. 1.

Q. E. D.

31 P. To draw a straight line through a given point (A), parallel to a given straight line (BC).

In BC take any point D, and join DA (a); make the angle DAE equal to the angle ADC (b); produce the straight line EA to F (c).

Because the alternate angles EAD and

ADC are equal (d), the straight line EF is B

D

parallel to the given straight line BC; and it is drawn through the given point A, which was to be done.

Recite (a) pos. 1;

(c) pos. 2;

(b) p. 23;

(d) p. 27.

32 Th. If any side (BC) of a triangle be produced, the exterior angle (ACD) is equal to the two interior angles (A and B); and the three interior angles of every triangle are equal to two right angles.

Argument. Through C draw CE parallel to AB (a). Then because AC meets the parallels AB, CE, the alternate angles A and ACE are equal (b).

Also, because BD meets the parallels AB, CE, the exterior DCE equals the interior B (c).

D

Therefore, the interior angles A and B equal the angles ACE and DCE, that is, the exterior angle ACD.

Again, to these equals add the angle ACB (d); therefore, the three interior angles A, B and ACB are equal to the two ACD, ACB, that is, to two right angles (e).

Wherefore, if any side, &c.

Recite (a) p. 31;

(d) ax. 2;

Q. E. D.

(b) p. 27;

(c) p. 29;

(e) p. 13.

Cor. 1. All the interior angles of any rectilineal figure and four right angles, are equal to twice as many right angles as the figure has sides. For, about a point within the figure, as many triangles may be formed as the figure has sides, each of whose angles shall equal two right angles; and the A angles about the point are equal to four right angles.

Cor. 2. All the exterior angles of any rectilineal figure are equal to four right angles; for such figure may be reduced to a mere point; about which there cannot be more than four D right angles.

B

33 Th. The straight lines (AC, BD) which join the same ends of two equal and parallel straight lines (AB, CD, are themselves equal and parallel.

Join the alternate ends BC (a); then since the equal and parallel lines AB, CD are met by BC, the alternate angles ABC, BCD are equal (b); and because AB equals CD, and BC is common; therefore the bases AC, BD are equal (c); also the alternate angles ACB, CBD (b); therefore also AC is parallel to BD.

Wherefore, the straight lines which join, &c.

Recite (a) pos 1; (b) p. 27; (e) p. 4.

Q. E. D.

B

Definition. A parallelogram is a quadrilateral figure whose opposite sides are parallel.

34 Th. The opposite sides and angles of a parallelogram are equal, and the diameter bisects it (a).

Because ABCD is a parallelogram, its opposite sides are parallel (b); and because BC joins opposite angles (c) it meets the parallels AB, CD, and also the parallels AC, BD, and makes the two angles at B alternately equal to the two at C (d); so that the whole angles ABD, ACD are equal.

Again, in the triangles ABC, DCB, the side BC is common; and their angles at B and C are alternately equal; therefore the remaining angles at A and D are equal (e);

B

also the sides AB to CD and AC to BD (f); therefore each of the triangles ABC, BCD is half of the parallelogram, which is bisected by BC.

Wherefore, the opposite sides and angles, &c.,

Recite (a), p. 9, 10;

(d), p. 27;

(b), def. above;
(e), p. 32;

Q. E. D.
(c), def. 16;

(f), p. 26.

35 Th. Parallelograms upon the same base (BC), and between the same parallels (AF, BC), are equal to one another.

To make ABCD, BCFD, distinct parallelograms on the same base, AF must be greater than BC; and the figure ABCF will be a trapezoid (a).

1. If AF be the double of BC, the straight lines DB, DC will bisect AF, and also each of the parallelograms; and the triangle DBC will be the half of each (b); hence the parallelograms ABCD, BCFD will be equal (c).

B

2. But if AF be greater than twice BC, DE B will be the excess; and if less, DE will be a common segment of AD, EF. To each, therefore, add the excess, or from each subtract the segment; the sums, or differences, AE, DF, will be equal (d). In either case, therefore, the sides CD, DF equal the sides BA, AE; and the exterior angle CDF equals the interior and opposite angle BAE (e), and the triangles CDF, BAE are equal (f). From the trapezoid ABCF take each of these triangles; the remaining parallelograms ABCD, BCFE are equal (d). Wherefore, parallelograms, &c. Recite (a), def. 34;

C

Q. E. D.

(b), p. 34;

(d), ax. 2, 3;

(e), p. 32;

(c), ax. 6;
(f), p. 4.