Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

EXERCISE XXXIII.

a

1. Show that XXVIII. is a double proposition, and enunciate the two separate propositions of which it is made up.

2. Enunciate a proposition which shall have a triple hypothesis, and include both XXVII. and XXVIII.

3. Write a converse to the proposition just referred to which shall therefore have a triple conclusion.

4. If three straight lines placed so as to form the letter Z contain two equal angles, show that two of them are parallel.

5. Show that two straight lines drawn perpendicular to the same straight line are parallel.

6. If the angles of a quadrilateral be all right, show that it is a parallelogram.

71. PROPOSITION XXIX.—THEOREM. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite upon the same side, and likewise the two interior angles upon the same side together equal to two right angles. Let the straight line E F fall upon the parallel straight lines

AB, CD, intersecting them in G and H. Then the alternate angles AGH, GHD shall be equal, the exterior angle EGB shall be equal to the opposite interior

angle GHD upon the same side of EF, and the two interior angles BGH, GHD, upon the same side of

EF, shall be together equal to two right angles. For, if the angle AGH be not equal to the angle GHD, one is greater than the other; let AGH be the greater;

[ocr errors]
[ocr errors][merged small][merged small]

and add to each of these unequals the angle BGH; then AGH and BGH are greater than BGH and GHD. (Ax. 4) But AGH and BGH are equal to two right angles ; (1.13)

therefore BGH and GHD are less than two right angles ; but those straight lines which, with another falling upon

them, make the interior angles on the same side of it together less than two right angles, will meet if continually produced ;

(Ax. 12) therefore AB, CD, if produced far enough, will meet. But they never meet, since they are parallel : (Hyp.)

therefore the angle AGH is not unequal to GHD; that is, AGH is equal to its alternate angle GHD.

But the angle AGH is equal to the angle EGB; (I. 15) therefore, also, EGB is equal to its opposite interior angle GHD.

Add to each of these the angle BGH; then EGB and BGH are equal to BGH and GHD; (Ax. 2) but EGB and BGH are equal to two right angles ; (I. 13) therefore also the two interior angles BGH, GHD are together

equal to two right angles. Wherefore, if a straight line, &c.

Q. E. D.

EXERCISE XXXIV.

(In 2–6, Diagram II. is referred to; in 7—9, Diagram III.) 1. Between angles belonging to the figure of Art, 68 deduce as many equalities as you can by the application of (a) first part of XXIX., (6) second part of the same, and (c) third part.

2. If the angleş EFG, EHG are equal, and FH bisects each of them, show, by Prop. XXVII., that EF is parallel to GH, and FG to EH.

3. If ABC, ADC are equal angles, and BD bisects each of them, prove ABCD a parallelogram.

4. The angles BRL, RSM are equal; show that RL, SM are parallel.

5. If PY, QX, BD be all parallel, show that the triangles APY, AQX, ABD are equiangular to one another.

6. If EFGH be a parallelogram, show that the angles adjacent to FH in the two triangles EF GFH form two of equal angles. By further application of XXVI., show that those triangles are equi. angular and have opposite sides equal.

7. The angles at A and B are right; apply XXVIII, to show that AD is parallel to BC.

8. If EF, GH be parallel to PQ, show that the angles EFG and HGU are each equal to PQF. Deduce that EF, GH are parallel..

9. If EFUV, HGUV are parallelograms, show that EFGH is a parallelogram.

10. Let BC be a straight line; at B and C make equal angles CBA, BCD, on opposite sides of BC; then show that AB and CD are parallel.

72. PROPOSITION XXX.—THEOREM.

Straight lines which are parallel to the same straight line are

parallel to one another. Let the straight lines AB, CD be each of them parallel to EF. Then shall AB be also parallel to CD.

[merged small][merged small][merged small][merged small][ocr errors][merged small]

Let the straight line GHK cut AB, CD, EF.

Then, because GHK cuts the parallel straight lines AB, EF, therefore the angle AGH is equal to the alternate angle HKF;

(I. 29) also because GHK cuts the parallels CD, EF, therefore the exterior angle GHD is equal to the interior angle HKF.

(I. 29) And it was shown that AGH is equal to HKF;

therefore the angle AGH is equal to the angle GHD; and these are alternate angles. Therefore AB is parallel to CD.

(I. 27) Wherefore, straight lines which are, &c.

Q. E. D.

[ocr errors]

Note.- The case in which EF lies between AB and CD needs no demonstration; otherwise, within an angle contained by AB and CD, it would be possible for EF to lie without cutting, when produced without limit, either AB or CD.

73. PROPOSITION XXXI.-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 a straight line through the point A parallel to the straight line BC.

In the line BC take any point D, and join AD; at the point A, in the straight line AD, make the angle DAE equal to the angle ADC; and produce the straight line EA to F.

Then EF shall be parallel to BC.

E

A

B

F

[blocks in formation]

(I. 23)

Because AD meets the two straight lines EF, BC, and makes the alternate angles EAD, ADC equal; therefore EF is parallel to BC.

(I. 27) Wherefore, through the given point A has been drawn a straight line EAF parallel to the given straight line BC.

Q. E. F.

74. "PLAYFAIR'S" AXIOM OF PARALLELISM.

It may be inferred, from previous observations on Axiom 12, that it is desirable to replace it by an axiom of a simpler nature, if that be possible. Many attempts have been made to accomplish this; one of which has met with so much approval in our own time that it is worth while to become acquainted with it.

The axiom proposed may be expressed thus:

Two straight lines which intersect cannot both be parallel to a third.

From this the usual Ax. 12 may very easily be deduced; and it will further clear up the nature of this important part of our subject to study some such proof as the following one :

Given Playfair's" Axiom, to prove Euclid's "6 TwelfthAxiom.
Let AC, BD be two straight lines, and AB another which falls upon

them, and let the two interior angles CAB, ABD on the same side of it be

together less than two right angles. Then AC, BD will meet, if produced, towards C, D. Make the angle BAX equal to ABE;

(I. 23) 80 that AX is parallel to BD.

(I. 27) F

х Y

C

E

D Then DBA, BAX are equal to DBA, ABE; and therefore to two right angles.

(I. 13) But DBA, BAC are less than two right angles ; therefore BAX is greater than BAC; that is, AC falls within the angle BAX,

and therefore AF falls beyond XY from B.

Now XY, FC intersect in A, and XY is parallel to ED;

(Constr.) therefore FC meets ED.

(Playfair's Axiom) But AF has been proved to fall beyond XY, therefore FC meets BD on the same side of AB as C.

Q. E. D. It has been objected to Axiom 12 that its converse is proved as Prop. XVII. It is similarly true of Playfair's Axiom that its converse is a proposition of Euclid, namely XXX.

EXERCISE XXXV. 1. How many distinct propositions are contained in XXVII., XXVIII., and XXIX.? What relation subsists among them ?

2. Props. XXVII., XXVIII., and XXX. may be said to establish four tests of the parallelism of two given straight lines. State these tests.

3. What proposition may be stated thus : “Two straight lines which are both parallel to a third cannot intersect" ?

State the relation of this proposition to “ Playfair's" Axiom. 4. Point out a mode of solving XXXI. by successive applications of XII. and XI., supposing the given straight line unlimited in length.

5. Prove XXVIII., denoting the angles by letters, as in the figure of Art. 68. 6. Prove XXIX. similarly.

7. Similarly prove XXX.

« ΠροηγούμενηΣυνέχεια »