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SYMBOLS AND ABBREVIATIONS. 47. The following are some of the principal symbols and abbreviations that are commonly used in Colleges and Public Schools. These symbols and abbreviations are generally employed in the present book. They are recommended to the student for writing out the propositions and demonstrations on the blackboard or in exercise books, as their use will greatly shorten the work. + plus, or together with. adj....... adjacent. minus, or diminished by.

..alternate. x multiplied by.

...axiom. • divided by.

.construction. .. therefore.

.corollary. equals,

cyl. ......cylinder.

.definition. > is (or are) greater than.

ext.

.exterior. < is (or are) less than.

fig.

.figure. L angle.

hyp...... hypothesis. Zs angles.

int. .interior. A triangle.

opp... opposite. As triangles.

post...... postulate. o parallelogram.

prob.....problem. Os parallelograms.

prop. proposition. I perpendicular.

pt.. point. Is perpendiculars.

rt. .right. 11 parallel.

.similar. Ils parallels.

sq. .square. O circle.

st. .straight. Os circles.

sup....... supplementary. O ce circumference.

O ces circumferences. The words "join AB” are used as an abbreviation for “draw a straight line from A to B.”

The initial letters Q. E. D., placed at the end of a theorem, stand for the Latin words Quod erat Demonstrandum, meaning which was to be proved.

The letters Q. E. F., placed at the end of a problem, stand for Quod erat Faciendum, which was to be done.

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Book I.
REOTILINEAR FIGURES.

B

A

F

To prove

PERPENDICULAR AND OBLIQUE LINES.

Proposition 1. Theorem. 48. All straight angles are equal to one another. *

Hyp. Let AB, AC be the arms of a st. , whose vertex is A, and DE, DF the arms of another st. L whose vertex is D.

Z BAC = _ EDF.
Proof. Because the BAC is a st. L,

.:. BA and AC are in the same st. line BAC. (21) Because the 2 EDF is a st. L,

.:. ED and DF are in the same st. line EDF. (21) Now if the _ BAC be applied to the _ EDF so that the vertex A shall coincide with the vertex D and the arm AB with the arm DE, then the arm AC will coincide with the arm DF, because, if two st. lines have two points in common they will coincide throughout their whole length, and forin but one st. line.

(Ax. 11) ... Z BAC = _ EDF.

Q.E.D. 49. COR. 1. All right angles are equal to one another. + (21), also (Ax. 7).

50. COR. 2. The complements of equal angles are equal to each other, and the supplements of equal angles are equal to each other.

51. CoR. 3. At a given point in a given straight line only one perpendicular can be drawn to that line.

* The Elements of Plane Geometry, by Association for the Improvement of Geometric Teaching, p. 20a. + This is Euclid's 11th axiom.

Proposition 2. Theorem.

52. If one straight line meet another straight line, the sum of the two adjacent angles is equal to two right angles. Hyp. Let the st. line DC meet the

E

D st. line AB at C.

To prove

ACD + L DCB = 2 rt. Zs.

A

B Proof. From C draw CE at rt. Zs to AB.

Then L ACD = _ ACE + L ECD. Add / DCB to each. .. LACD +_ DCB = _ ACE + L ECD + _ DCB.

If equals be added to equals, the sums will be equal (Ax. 2). Again, Z ECB = _ ECD + L DCB. Add / ACE to each. ... Z ACE + Z ECB = _ ACE + L ECD + L DCB.

If equals be added to equals, the sums will be equal (Ax. 2). Hence

_ ACE + _ ECB = _ ACD + _ DCB. (Ax. 1)* But _ ACE+_ ECB = 2 rt. Zs.

(Cons.) ... ACD + ZDOB = 2 rt. Zs.

Q. E.D.

Sch. The angles ACD, DCB are supplementary adjacent angles.

(25) 53. COR. If one of the angles ACD, DCB is a right angle, the other is also a right angle.

The sum of all the angles on the same side of a straight line, at a common point, is equal to two right angles ; for their sum is equal to the sum of the two adjacent angles ACD, DOB.

* The student should quote every reference in full.

Proposition 3. Theorem.

54. Conversely, if the sum of two adjacent angles is equal to two right angles, their exterior sides are in one straight line.

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Hyp. Let LACD + _ DCB = 2 rt. Zs.
To prove that AC and CB are in one st. line.

Proof. If CB be not in the same st. line as AC, let CE be in the same st. line as AC. Then _ ACD + Z DCE = 2 rt. 28,

being sup. adj. Z8 (52). But / ACD + Z DCB = 2 rt. Zs.

(Hyp.) .. _ ACD + _ DCE = _ ACD + _ DCB. (Ax. 1) Take away the common / ACD. .: LDCE = DCB,

(Ax. 3) which is impossible (Ax. 8), unless CE coincides with CB.

... AC and CB are in one st. line. Q. E.D.

EXERCISES.

1. Find the number of degrees in an angle if it is three times its complement.

2. Find the number of degrees in an angle if its complement and supplement are together equal to 110°.

. Proposition 4. Theorem. 55. If two straight lines intersect each other, the vertical angles are equal.

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To prove

Hyp. Let the st. lines AB and CD cut at E.

2 AEC = _ BED. Proof. Z CEA + Z CEB = 2 rt. ZS,

being sup. adj. 48 (52).
AEC + Z AED = 2 rt. 28,

being sup. adj. 28 (52).
i. Z CEA + Z CEB = L AEC + 2 AED.

All rt. 28 are equal to one another (49).
Take away the common _ AEC.
..Z CEB = _ AED.

(Ax. 3) In the same way it may be proved that

Z AEC
< BED.

Q.E.D.

56. COR. 1. If two straight lines intersect each other, the four angles which they make at the point of intersection, are together equal to four right angles.

If one of the four angles is a right angle, the other three are right angles, and the lines are mutually perpendicular to each other.

57. Cor. 2. If any number of straight lines meet at a point, the sum of all the angles made by consecutive lines, in a plane, is equal to four right angles.

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