Sidebilder
PDF
ePub

the angle B to the angle E, therefore the side AC is equal to DF, and the angle A to the angle D (d).

[2.] Let the equal sides be BA and ED, opposite to the equal angles C and F, then the side BC is equal to EF.

For if it be possible, let one of them BC be the greater; make BH equal to EF (a), and join AH. Then, because in the triangles ABH and DEF, the sides BA and BH

да

are equal to ED and EF (6), and the angle B equal to the angle E (c), therefore the angle AHB is equal to the angle F (d); but the angle C is also equal to the angle F (c), therefore the angle AHB is equal to the angle C (e), that is, the external angle AHB of the triangle AHC is equal to the internal opposite angle C, which is impossible (f); therefore neither of the sides BC and EF is greater than the other, and therefore they are equal. And because in the triangles ABC and DEF the sides BA and BC are equal to ED and EF, and the angle B to the angle E, therefore the side AC is equal to DF, and the angle A to the angle D (d).

SCHOLIUM. It is evident that the triangles are themselves equal.

COROLLARY 1. If a straight line (AB) be drawn from the vertex of an isosceles triangle (ACD) perpendicular to the base, it will bisect the base, and also the angle (CAD) opposite to the base.

For in the triangles ABC and ABD the angles C and ABC are equal to the angles D and ABD (u), and the side AB, opposite the equal angles C and D, is common to both; therefore the angle CAB is equal to the angle DAB, and the side CB to BD (b); and therefore the base CD is bisected, and also the angle CAD opposite to the base.

(a) Hypoth. (b) I. 26.

COROLLARY 2. It is evident that a straight line which bisects the angle opposite to the base of an isosceles triangle, bisects also the base, and is perpendicular to it; and that a straight line drawn from the vertex, bisecting the base, is perpendicular to it, and bisects the opposite angle.

PROPOSITION XXVII.

THEOREM.-If a straight line (EF) intersect two other straight lines (AB and CD), both in the same plane, and form alternate angles equal to each other (AEF to EFD or BEF to EFC), these two straight lines shall be parallel.

DEMONSTRATION. For if possible let AB and CD not be parallel, but meet when produced on the side BD in some point, as G. Then, in the triangle EGF, the external angle AEF is greater than the in

ternal opposite angle EFG (a); but it is also equal to it (b), which is absurd; therefore the lines AB and CD do not meet on the side BD, and in like manner it can be proved that they do not meet on the side AC. Since, then, the lines AB and CD, when produced on either side, do not meet, they are parallel (c).

(a) I. 16.
(b) Hypoth.
(c) Def. 7.

SCHOLIUM. The condition that both the straight lines AB and CD shall be in the same plane is necessary to be introduced in the enunciation of this and the two following propositions; for it would be possible for two straight lines to accord with the remainder of the hypothesis, and yet not to be parallel if they were not in the same plane.

PROPOSITION XXVIII A.

THEOREM.If a straight line (EF) intersect two other straight lines (AB and CD), both in the same plane, and form an external angle equal to the internal and opposite angle upon the same side of the line (EGB to GHD), the two straight lines shall be parallel.

DEMONSTRATION. Because the angle EGB is equal to the angle GHD (a), and the angle EGB is also equal to the vertical angle AGH (6), therefore AGH is equal to GHD (c); but they are the alternate angles, therefore ÀВ is parallel to CD (d).

H

(a) Hypoth.
(b) I. 15.
(c) Ax. 1.

(d) I. 27.

PROPOSITION XXVIII B.

THEOREM.-If a straight line (EF) intersect two other straight lines (AB and CD)), both in the same plane, and form internal angles at the same side (BGH and GHD) equal to two right angles, the two straight lines shall be parallel.

H

(a) Hypoth.
(b) I. 13.
(c) Ax. 1.
(d) I. 27.

DEMONSTRATION. Because the angles BGH and GHD are equal to two right angles (a), and the angles AGH and BGH are also equal to two right angles (6), therefore the angles BGH and GHD are equal to the angles AGH and BGH (c); take away from both the common angle BGH, and

the remaining angles GHD and AGH are equal; but they are the alternate angles, therefore AB is parallel to CD (d).

SCHOLIUM. The twenty-eighth proposition of Euclid really consists of two distinct propositions, which are here distinguished and separately demonstrated.

PROPOSITION XXIX.

THEOREM.-If a straight line (EF) intersect two parallel straight lines (AB and CD), [1] it forms the alternate angles equal to one another (AGH to GHD), [2] and the external angle equal to the internal and opposite angle upon the same side (EGB to GHD), [3] and also the two internal angles on the same side (BGH and GHD), together equal to two right angles.

DEMONSTRATION. [1.] For if AGH is not equal to GHD, let KG be drawn, making the angle KGH equal to the internal opposite angle GHD (a), and produce KG to L; then the line KL will be parallel to CD (6); but AB is also parallel to CD (c); therefore through the same point G two straight lines have been drawn parallel to the same straight line, which is impossible (d); the alternate angles AGH and GHD are therefore equal.

H

(a) I. 23.
(b) I. 27.
(c) Hypoth.
(d) Ax. 12.
I. 15.

Ax. 1.

(g) Ax. 2.

[2.] The external angle EGB is equal to the internal and opposite angle on the same side GHD; for the angle EGB is equal to the vertical angle AGH (e), and the angle AGH has been proved to be equal to the alternate angle GHD, therefore EGB is equal to GHD (ƒ).

I. 13.

[3.] The two internal angles on the same side BGH and GHD are together equal to two right angles. For since EGB is equal to GHD, add BGH to both, therefore EGB and BGH are equal to GHD and BGH (g); but EGB and BGH are equal to two right angles (h), therefore also BGH and GHD are equal to two right angles.

SCHOLIA. 1. Euclid's twenty-ninth proposition really consists of three distinct propositions, which are here separated; [1] is the converse of proposition xxvii., [2] is the converse of proposition xxviii A, and [3] is the converse of proposition xxviii B.

2. The demonstration here given is similar to that of Playfair, and differs slightly from that given by Euclid, being proved by means of the axiom, "That through the same point two straight lines cannot be drawn parallel

to the same straight line," instead of being proved by the twelfth axiom as given by Simson, which, being the converse of another proposition (I. 17), requires itself to be demonstrated, which may be done in the following

manner.

THEOREM. If a straight line (EF) meets two straight lines (AB and CD), so as to make the two internal angles on the same side (BGH and GHD) together less than two right angles, these straight lines (AB and CD) being continually produced, shall at length meet upon that side on which are the angles which are less than two right angles.

DEMONSTRATION. For if the two straight lines AB and CD do not meet when continually produced, they are parallel (a); and if GL be drawn, making the angles LGH and GHD together equal to two right angles, it will be parallel to CD (b). That is, through the point G the two straight lines GL and AB have been drawn both parallel to the line CD, which is impossible (c); therefore the straight lines AB and CD are not parallel, but shall at length meet if continually produced.

COROLLARY. The parts of all perpendiculars to two parallel straight lines, intercepted between them, are equal.

Let EF and GH be perpendicular to the two parallel straight lines AB and CD; and join EH. In the triangles FEH and GHE the angles FEH and GHE are equal, being alternate (a), and the angles EHF and HEG are equal, being alternate (a), and the side EH is common to both; therefore the remaining sides are respectively equal to each other (b), the side EF to the side GH.

E

(a) Def. 7.
(b) I. 28 B
(c) Ax. 12.

E

IG

H

(a) I. 29.
(b) I. 26.

PROPOSITION XXX.

THEOREM.-If two straight lines (AB and CD) be parallel to the same straight line (EF), they are parallel to each other.

DEMONSTRATION. Let the straight line GHK intersect AB, EF, and CD. Then the angle AGK is equal to the alternate angle GHF (a), and the external angle GHF is equal to the internal opposite angle GKD (a), therefore the angle AGK is equal to the angle GKD (6); and because they are alternate angles, and are equal, therefore AB is parallel to CD (c).

E

C

B

H

(a) I. 29.
(b) Ax. 1.

I. 27.

PROPOSITION XXXI.

PROBLEM.-Through a given point (A), to draw a straight line parallel to a given straight line (BC).

SOLUTION. In the line BC take any point D, join AD, and at the point A form the angle DAE with the straight line AD, equal to the angle ADC, and on the opposite side of the line AD (a), and produce EA to F; then the line EF is parallel to BC.

(a) I. 23.

Solution.

(c) I. 27.

DEMONSTRATION. For the straight line AD, intersecting the lines EF and BC, forms the alternate angles DAE and ADC equal (6), therefore the lines EF and BC are parallel (c).

PROPOSITION XXXII A.

THEOREM.-If any side (BC) of a triangle (ABC) be produced, the external angle (ACD) is equal to the sum of the two internal and opposite angles (A and B).

CONSTRUCTION. Through the point C draw CE parallel to the straight line AB (a).

DEMONSTRATION. Because BA and CE are parallel (6), the angle ACE is equal to the alternate angle A (c), and the angle ECD is equal to the internal angle B (c), therefore the whole external angle ACD is equal to the two internal angles A and B (d).

(a) I. 31.

(b) Const.

(c) I. 29.

(d) Ax. 2.

SCHOLIUM. Euclid's thirty-second proposition really consists of two distinct theorems, which are here separately distinguished.

PROPOSITION XXXII B.

THEOREM.-If any three angles are the internal angles of a

triangle (ABC), they are together equal to

two right angles.

DEMONSTRATION. Produce BC to D. Then the angles ACB and ACD are equal to two right angles (a), they are also equal to the three internal angles A, B, and ACB (6); therefore the three internal angles of the triangle ABC are together equal to two right angles (c).

(a) I. 13.
(b) I. 32 A.
(c) Ax. 1.

« ForrigeFortsett »