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7 Th. Upon the same base (AB), and on one side of it, two triangles cannot have their sides (AC, AD) equal, which terminate in one extremity of it, and likewise their sides (BC, BD), which terminate in the other extremity.
Join CD: then, in the case in which the vertex of each triangle is without the other; because AC and AĎ are equal, the angles ACD. ADC, are also equal (a). But the angle BCD is less than the angle ACD, therefore less than ADC, and still less than BDC. Again, because CB is equal to DB, the angles BCD, BDC, are equal: but BCD has been proved to be much less than BDC; both equal and less is impossibie.
And, in the case of one of the vertices, D, being within the other triangle ACB, produce AC, AD, to E, F; therefore, because ACD is an isosceles triangle, the angles ECD, FDC, beyond the base, are equal (6); but the angle BCD is less than the angle ECD, therefore less than FDC, and still less than BDC. Again, because BC and BD are equal, the angles BDC, BCD, are also equal : but BCD has been proved to be less than BDC, which is impossible.
The case in which the vertex of one triangle is upon the side of another, needs no demonstration. Therefore, upon the same base, &c.
Q. E. D. Recite (a), prop. 5, and (b), prop. 5.
8 Th. If two triangles have two sides (AB, AC) of the one, equal to two sides (DE, DF) of the other, each to each, and have likewise their bases (BC, EF) equal ; the angles (A, D) contained by the corresponding sides, shall be equal to one another.
Demonstration. Apply the triangle ABC to DEF; so that the point B fall on E, and the base BC fall upon its equal EF; then the point C shall coincide with F, and the sides BA, AC, fall on ED, DF, each upon each; for, if the bases coincide, but the sides fall in another direction, as on G, then upon one side of the base EF, there can be two triangles, with equal sides, which terminate in either extremity of the base; which is impossible (a). Wherefore, if the bases
coincide, the sides shall also coincide; and, consequently, the angles BAC, EDF, contained by those sides, shall be equal. Q. E. D.
Recite (a) prop. 7.
9 P. To bisect a given rectilineal angle (BAC); that is, to divide it into two equal angles.
Constr. In AB take any point D; make AE equal to AD (a); join DE (1), and upon it describe an equilateral triangle DEF (c); join AF: the straight line AF bisects the angle BAC.
Dem. Because AD is made equal to AE, and; AF is cominon to the two triangles DAF, EAF,
E and the bases DF, EF, were made equal: therefore the angle DAF is equal to EAF (d); that is, the rectilineal angle BAC is bisected by the straight line AF; which was to be done.
Recite (a), prop. 3; (6), post. 1; (c), prop. 1; (d), prop. 8.
10. P. To bisect a given finite straight line (AB); that is, to divide it into two equal parts.
Constr. Upon AB describe an equilateral triangle ABC (a), and bisect the angle ACB by the straight line CD (6).
Argument. AC is made equal to BC, the angle ACD to BCD, and CD is common to the two triangles ACD, BCD: therefore, the bases AD and BĎ are equal, and AB is cut into two equal parts in the point D. Which was to be done.
Recite (a), prop. 1; (b), prop. 9.
11 P. To draw a straight line (CF), at right angles to a given straight line (AB), from a given point (C), in the
Construction. In AC take any point D, and make CE equal to CD (a); upon DE describe the equilateral triangle DFE (6); join FC (e);
Argument. The sides CD, CF, are equal to the sides CE, CF; and the bases DF, EF, are also equal: therefore the included angles at C, being equal (d), and adjacent, the straight line FC is drawn from the given point C, at ĀD right angles to AB (e); which was to be done. Recite (a), prop. 3;
(6), prop 1; (c), post. 1; (d), prop. 8;
(e), def. 10.
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, (b): A therefore the lines ABC, ABD, which have the cominon segment AB, are not both straight lines.
Q. E. D. 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.
Argument. The triangles CHF, CHG have CH common, HF equal to HG, and
GB the bases CF, CG are equal radii; therefore, 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, (6) prop. 10, (c) post. 1, (d) prop. E, (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.
Construction. If the adjacent angles ABC, ABD, are equal to one another, they are right angles (a); if unequal, from the point B draw BE at right angles to CD (6).
Argument. The two right angles EBD, EBC are equal to the three EBD, EBA and ABC.
Again, the two angles ABD, ABC are equal to the three EBD, EBA, ABC; therefore ABC, ABD are equal to EBD, EBC (c), which were made equal to two right angles.
Therefore, the angles, &c. Q. E. D.
c), ax. 1.
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 augles ABC, ABE, are equal to two riglit 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.
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 eqnal to two right angles; and the straight line DE makes with AB, the angles DEA, DEB equal to two right angles (a); from these equals take the cominon angle AED: therefore, the remainders AEC, BED are equal to one another (6).
In the same manner, it can be demonstrated that the angle AED is equal to the angle BEC. Therefore, if two straight lines, &c.
Q. E. D. 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 inade 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.
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.
(6), pos. 1;
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 two of the angles of ABC are less than two right angles.
Q. E. D. Recite (a), pos. 2; (6), p. 16; (c), ax. 4; (d), p. 13.
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 (6).
Argument. The exterior angle ADB, or its equal ABD (c), exceeds the interior C (d); much inore does ABC exceed C. Therefore, in every side, &c.
Q. E. D. Recite (a), p. 3; (b), pos. 1;
(), 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).
Q. E. D. Recite (a), p. 5; (), p. 18.