Cor. Hence two straight lines cannot have a common segment: for if ABC, ABD, have the segment AB common, they cannot both be straight lines. Draw BE at right angles to AB (a): then if ABC be a straight line, EBC is a right angle; and if ABD be a straight line, EBD is a right angle; and so two right angles are unequal, which is impossible, (6): A therefore the lines ABC, ABD, which have

E

D

the cominon segment AB, are not both straight lines.

Recite (a), p. 11; (b), ax. 10.

12 P. To draw a straight line (CH), perpendicular to a given straight line (AB), of sufficient length, from a given point (C) on one side of it.

Construction. Take any point D, on the other side of AB; and from the centre C, at the distance CD, describe the arc EGF (a), cutting AB in F, G; bisect FG in H (6), and join CF, CH, CG, (c); CH is perpendicular to AB.

E

Argument. The triangles CHF, CHG

have CH common, HF equal to HG, and A F

the bases CF, CG are equal radii; there

D

fore, the angles CHF, CHG, are equal (d),

and being adjacent, they are right angles (e), and CH drawn from the point C, is consequently perpendicular to AB: which was to be done. Recite (a) post. 3, (b) prop. 10, (c) post. 1, (d) prop. 8, (e) def. 10.

13 Th. The angles (ABC, ABD), made by one straight line (AB) with another (CD), upon one side of it, are either two right angles, or are equal to two right angles.

14 Th. If, at a point (B), in a straight line (AB), two other straight lines (CB, DB) meet from opposite sides, making the adjacent angles equal to two right angles, these two lines shall be one continued straight line.

Argument. If CBD be not a straight line, make CBE such: therefore, since AB makes angles with the straight line CBE, on one side of it, the angles ABC, ABE, are equal to two right angles (a): but the angles ABC, ABD, are equal to two right angles. Take away the common angle ABC: therefore the remainders ABE, ABD, are equal (6); the less to the greater, which is impossible. Wherefore, if at a point, &c.

E

B

D

Q. E. D.

Recite (a), prop. 13; (b), ax. 3.

15 Th. If two straight lines (AB, CD), cut one another, the opposite, or vertical angles (AEC, BED) and (BEC, AED) shall be equal to one another.

Argument. Since the straight line AE makes with CD, the angles AEC, AED equal C

to two right angles; and the straight line DE makes with AB, the angles DEA, DEB equal A to two right angles (a); from these equals take the common angle AED: therefore, the re

mainders AEC, BED are equal to one another (b).

is equal to the angle BEC.

Therefore, if two straight lines, &c.

Q. E. D.

In the same manner, it can be demonstrated that the angle AED

Recite (a), prop. 13; (b), ax. 3.

Cor. 1. Hence if two straight lines cut one another, the angles made

at the sectional point are equal to four right angles.

Cor. 2. Consequently, all the angles made about a point, by any

number of cutting lines, are equal to four right angles.

16 Th. If one side (BC) of a triangle (ABC) be produced, the exterior angle (ACD) is greater than either of the interior opposite angles (at A, or B).

Constr. Bisect AC in E (a); join BE (6) and produce it to F (c); make EF equal to EB; join FC, and produce AC to G.

Argument. The triangles AEB, CEF are equal; having two sides EA, EB in the one equal to two sides EC, EF in the other; also their vertical angles at E being equal (d); therefore, their angles A and ECF are equal. But the exterior angle ECD is greater than ECF, or A.

B

E

D

In like manner, if BC be bisected, it may be proved that the angle

BCG, or its equal ACD, is greater than the angle ABC.

Therefore, if one side of a triangle, &c.

Q. E. D.

Recite (a), p. 10;

(b), pos. 1;

(c), pos. 2;

(d), p. 4, 15.

17 Th. Any two any angles of triangle (ABC) are together less than two right angles.

Argument. Produce the side BC to D (a). Then since the interior angle B is less than the exterior and opposite angle ACD (6), to each add ACB; then ACB and B are less than ACB and ACD (c): but these latter two are equal to two right angles (d); therefore, ACB and B are less than two right angles.

In the same way, it may be proved that any f two of the angles of ABC are less than two right angles.

A

C

Q. E. D.

Recite (a), pos. 2; (b), p. 16; (c), ax. 4; (d), p. 13.

D

18 Th. In every triangle (ABC) the greater angle (B) is opposite to the greater side (AC).

Constr. Since the side AB is less than AC, cut off a part AD equal to AB (a); join BD (b). Argument. The exterior angle ADB, or its equal ABD (c), exceeds the interior C (d); much more does ABC exceed C.

Therefore, in every side, &c.

A

D

B

Q. E. D.

Recite (a), p. 3;

(b), pos. 1;

(c), p. 5;

(d), p. 16.

19 Th. The greater angle of every triangle is subtended by the greater side, and the less by the less.

Argument. If in the triangle ABC, the angle B be greater than C, the side AC will exceed the side AB: for if not, it must be either equal to it, or less: equal it is not, because B is not equal to C (a); neither is it less, because B is not less than C. It remains, therefore, that AC is greater than AB, because it subtends a greater angle (b).

Recite (a), p. 5; (6), p. 18.

A

Q. E. D.

20 Th. Any one side of a triangle is less than the sum of the other two.

Constr. In the triangle ABC, produce BA so as to make AD equal to AC: join CD (a). Argument. The angles ACD, ADC are equal, being opposite to equal sides (a); but either of them is less than BCD; and the less side subtends the less angle (b); therefore BC is less than BD, which is the sum of BA and AC.

In this manner it may be proved that either side is less than the sum of the other two.

Recite (a), pos. 1; (6), p. 5, 19.

Q. E. D.

21 Th. If from a point (D) within a triangle (ABC), two straight lines (DB, DC) be drawn to the ends of the base; these lines shall be less than the two sides (AB, AC), but they shall contain a greater angle.

Argument 1. Produce BD to E (6). BE is less than the sum of BA, AE (6); add EC to both. Then the sum of BE and EC is less than the sum of BA and AC (c). Again CD is less than the sum of CE and ED (6); to both add DB; then the sum of CD, DB is less than the sum of CE, EB, and still less than the sum

A

E

D

of CA, AB (c).

2. The angle BEC is exterior of the triangle BAE, and interior of the triangle CED; it is therefore less than the exterior BDC, and greater than the interior BAC: BDC is therefore greater than BAC (d). Q. E. D.

Wherefore, if from a point, &c.

Recite (a), pos. 2; (6), p. 20; (c), ax. 4; (d), p. 16.

22 P. To make a triangle of which the sides shall be equal to three given straight lines (A, B, C); but any one of them must be less than the sum of the other two.

Argument. FK, FD are equal radii (c); and FD equals A; therefore FK equals A (d): also GK, GH are equal radii; and GH equals C; therefore GK equals C. Therefore FK, FG, GK are equal to A, B, C, each to each; and the thing is done which was required. Recite (a), p. 3; (b), pos. 3; (c), def. 15; (d), ax. 1.

23 P. At a given point (A) in a given straight line (AB) to make an angle equal to a given rectilineal angle (C).

Constr. In the lines forming the angle C join any two points D, E (a); make the triangle AFG of sides equal to CD, CE, DE, each to each (b).

Argument. Since FA, AG are made equal to DC, CE, and FG to DE; therefore, at the point A, in the straight line AB, an angle is made equal to the angle C (c); which was to D be done.

Recite (a), pos. 1; (6), p. 22; (c) p. 8.

0

A

E

F
B/

G

24 Th. If two sides (AB, AC) of one triangle, be equal to two sides (DE, DF) of another, while the contained angles (A and D) are unequal; the base (EF) opposite the less angle (D) is less than the base (BC) opposite the greater angle (A).

Constr. Let DF be not less than DE: then, at the point D. in the straight line ED, make the angle EDG equal to A (a); make DG equal to AC, or DF (b); join EG, GF (c).

D

Dem. The two sides BA, AC being severally equal to the two ED, DG, and the angle A to EDG, the bases BC, EG B

are therefore equal (d). But, in the trian

gle EFG, EF is less than EG; for it subtends a less angle-that is, since DFG is isosceles (d), the angles DFG, DGF are equal; but EGF is less, and EFG is greater than either: therefore EF is less than BC.

Wherefore, if two sides, &c.

Q. E. D.

Recite (a), p. 23; (b), cor. p. 3; (c), post. 1; (d), p. 5.

25 Th. If two sides of one triangle be severally equal to two sides of another, but the bases unequal; the angle opposite the greater base is greater than the other.