« ΠροηγούμενηΣυνέχεια »
(ii) Again, because the angle AGH is equal to the vertically opposite angle EGB;
I. 15. and because the angle AGH is equal to the angle GHD;
Proved. therefore the exterior angle EGB is equal to the interior
opposite angle GHD.
(iii) Lastly, the angle EGB is equal to the angle GHD;
Proved. add to each the angle BGH; then the angles EGB, BGH are together equal to the angles
BGH, GHD. But the adjacent angles EGB, BGH are together equal to
two right angles : therefore also the two interior angles BGH, GHD are together equal to two right angles.
EXERCISES ON PROPOSITIONS 27, 28, 29. 1. Two straight lines AB, CD bisect one another at O: shew that the straight lines joining AC and BD are parallel. (1. 27.]
2. Straight lines which are perpendicular to the same straight line are parallel to one another.
[I. 27 or 1. 28.] 3. If a straight line meet two or more parallel straight lines, and is perpendicular to one of them, it is also perpendicular to all the others.
[1. 29.] 4. If two straight lines are parallel to two other straight lines, each to each, then the angles contained by the first pair are equal respectively to the angles contained by the second pair.
NOTE ON THE TWELFTH AXIOM.
Euclid's twelfth Axiom is unsatisfactory as the basis of a theory of parallel straight lines. It cannot be regarded as either simple or self-evident, and it therefore falls short of the essential characteristics of an axiom : nor is the difficulty entirely removed by considering it as a corollary to Proposition 17, of which it is the
Of the many substitutes which have been proposed, we need only notice the following :
Axiom. Two intersecting straight lines cannot be both parallel to a third straight line.
This statement is known as Playfair's Axiom; and though it is not altogether free from objection, it is no doubt simpler and more fundamental than that employed by Euclid, and more readily admitted without proof.
Propositions 27 and 28 having been proved in the usual way, the first part of Proposition 29 is then given thus.
PROPOSITION 29. [ALTERNATIVE PROOF.] If a straight line fall on two parallel straight lines, then it shall make the alternate angles equal.
Let the straight line EF meet the two parallel straight lines AB, CD at G and H. Then shall the alternate angles AGH, GHD
E be equal.
B For if the angle AGH is not equal to the
angle GHD: at G in the straight line HG make the angle HGP equal to the angle GHD, and
с H alternate to it.
I. 23. Then PG and CD are parallei.
F But AB and CD are parallel : Hyp. therefore the two intersecting straight lines AG, PG are both parallel to CD: which is impossible.
Playfair's Axiom Therefore the angle AGH is not unequal to the angle GHD;
that is, the alternate angles AGH, GHD are equal. Q.E.D. The second and third parts of the Proposition may then be deduced as in the text; and Euclid's Axiom 12 follows as a Corollary.
PROPOSITION 30. THEOREM. Straight lines which are parallel to the same straight line are parallel to one another.
Let the straight lines AB, CD be each parallel to the straight line PQ.
Then shall AB and CD be parallel to one another. Construction. Draw any straight line EF cutting AB, CD, and PQ in the points G, H, and K.
Proof. Then because AB and PQ are parallel, and EF meets them, therefore the angle AGK is equal to the alternate angle GKQ
I. 29. And because CD and PQ are parallel, and EF meets them, therefore the exterior angle GHD is equal to the interior opposite angle GKQ.
I. 29. Therefore the angle AGH is equal to the angle GHD;
and these are alternate angles ; therefore AB and CD are parallel.
I. 27. Q.E.D.
Note. If PQ lies between AB and CD, the Proposition may be established in a similar manner, though in this case it scarcely needs proof; for it is inconceivable that two straight lines, which do not meet an intermediate straight line, should meet one another.
The truth of this Proposition may be readily deduced from Playfair's Axiom, of which it is the converse.
For if AB and CD were not parallel, they would meet when produced. Then there would be two intersecting straight lines both parallel to a third straight line : which is impossible.
Therefore AB and CD never meet; that is, they are parallel.
PROPOSITION 31. PROBLEM. To draw a straight line through a given point parallel to a given straight line.
Let A be the given point, and BC the given straight line. It is required to draw through A a straight line parallel to BC.
Construction. In BC take any point D; and join AD. At the point A in DA, make the angle DAE equal to the angle ADC, and alternate to it,
I. 23. and produce EA to F.
Then shall EF be parallel to BC. Proof. Because the straight line AD, meeting the two straight lines EF, BC, makes the alternate angles EAD, ADC equal;
Constr. therefore EF is parallel to BC ;
I. 27. and it has been drawn through the given point A.
1. Any straight line drawn parallel to the base of an isosceles triangle makes equal angles with the sides.
2. If from any point in the bisector of an angle a straight line is drawn parallel to either arm of the angle, the triangle thus formed is isosceles.
3. From a given point draw a straight line that shall make with a given straight line an angle equal to a given angle.
4. From X, a point in the base BC of an isosceles triangle ABC, a straight line is drawn at right angles to the base, cutting AB in Y, and CA produced in Z: shew the triangle AYZ is isosceles.
5. If the straight line which bisects an exterior angle of a triangle įs parallel to the opposite side, shew that the triangle is isosceles,
PROPOSITION 32. THEOREM. If a side of a triangle be produced, then the exterior angle shall be equal to the sum of the two interior opposite angles; also the three interior angles of a triangle are together equal to two right angles.
Let ABC be a triangle, and let one of its sides BC be produced to D. Then (i) the exterior angle ACD shall be equal to the sum of the
two interior opposite angles CAB, ABC ; (ii) the three interior angles ABC, BCA, CAB shall be
together equal to two right angles. Construction. Through C draw CE parallel to BA. I. 31. Proof. (i) Then because BA and CE are parallel, and AC
meets them, therefore the angle ACE is equal to the alternate angle CAB.
I. 29. Again, because BA and CE are parallel, and BD meets them, therefore the exterior angle ECD is equal to the interior opposite angle ABC.
I. 29. Therefore the whole exterior angle ACD is equal to the sum of the two interior opposite angles CAB, ABC.
(ii) Again, since the angle ACD is equal to the sum of the angles CAB, ABC;
Proved. io each of these equals add the angle BCA: then the angles BCA, ACD are together equal to the three
angles BCA, CAB, ABC. But the adjacent angles BCA, ACD are together equal to two right angles.
1. 13. Therefore also the angles BCA, CAB, ABC are together equal to two right angles.