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COR.-Hence it may be shown that two st. lines cannot have a common segment.

EXP. |1| If possible, let the segment

AB be common to the
lines ABC and ABD.

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E

DEM. 1 Hyp.

2 Def. 10.

3 Hyp.

4 Def. 10.

5 D.2,4,Ax.1

6 Recap.

making the

Ls with AB A

rt. angles.

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AB and BC make one st. line ABC, and

LABE is a rt. angle;

.. LABE= [EBC:

Again,

AB and BD form one line ABD,

and ABE is a rt. angle;

LABEEBD:

Wherefore EBD/EBC; which is impossible.

Therefore two st. lines cannot have a common segment.

Q.E.D.

SCHOLIUM.-1. This corollary should be ranked among the axioms; for it is assumed in Prop. 4, where it is taken for granted that if certain lines, placed on one another, coincide for any portion of their length, they must coincide throughout. It is also assumed in Prop. 8.

2. When the given point is at the extremity of a given line, the line from that extremity should be produced, and the rt. angle be then constructed:

3. Or the following method may be adopted :-Let AB be the given line, and B the extremity from which the perpendicular is to be raised. Take any point G above the line AB, and with radius GB describe a circle cutting AB in H; join HG and produce it to M; or from H set the same radius GB on HK, KL, and LM; M is the point from which if a line be drawn to A

B, that line MB will be perpendicular

to AB. The radius equals the chord

of 60°; three times 60° equal 180°;

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and 180° is the semicircle: and by P. 31, bk. iii., the angle HBM in a semicircle is a rt. angle.

USE AND APPLICATION.-1. By this proposition the Square is constructed, an instrument employed for ascertaining the perpendicular to a horizontal line, and for all purposes for which right angles are needed.

2. On a given line AB, to describe an isosceles triangle of which the perpendicular height CD, is equal to the base A B.

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CONS. 1 P. 10, 11, & 3 Bisect AB in D, and make DC perpen. and equal to

2 Pst. 1.
3 Sol.

DEM. 1 C. 1.

2 P. 4.
3 C. 1.

AB;

join AC and BC;

the figure ABC is the isosc. A required.

::AD=DB, DC is common, and ▲ ADC = 4 BDC; .. AC = BC.

And the perp. DC has been made equal to AB.

Q. E. F.

PROP. 12.-PROB.

To draw a perpendicular to a given st. line of unlimited length from a given point without it.

SOLUTION.-Pst. 3. A circle may be drawn from any centre at any distance from that centre.

P. 10. To bisect a given st. line.

Pst. 1. A line may be drawn from one point to another.

DEMONSTRATION.-Def. 15. The radii of the same circle are all equal. P. 8. If two triangles have two sides in one equal to two sides in the other, and the base equal to the base, the angles contained by the two pair of equal sides are equal.

Def. 10. A st. line at rt. angles to another st. line, is perpendicular to it.

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EXP. 1 Data. 2 Quæs.

CONS. 1 Assum.

2 Pst. 3.

K

Given the st. line AB, and the C out of it; required from C a perpendicular to AB.

Take D on the other side of AB;

from C with rad. CD draw the arc EGF, cutting AB in F and G;

3 P.10, Pst.1 bisect FG in H, and join C and H, C and F, and C and G;

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SCHOLIUM.-1. The properties of the circle form the subject of the third book, but in the construction for the 12th Prop., the Lemma is borrowed from bk. iii., that the circle will intersect the line in two points.

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In the triangles NCO, NMO, the three sides of the one are equal to the three sides of the other, and by P. 8, the angle CNO equals the angle MNO thus in the triangles NDC, NMD, two sides and the included angle of one are equal to two sides and the included angle of the other:therefore by P. 4, ang. ADC equals ang. ADM, and they are adjacent angles; hence CD is perpendicular to A B.

USE AND APPLICATION.-1. In practice, the problem will be solved by drawing, as in the figure to P. 12, the arc FDE, and from the points F and G, with equal radii, describing arcs intersecting in K; by joining CK, the perpendicular to AB will be drawn.

2. This problem is indispensable to all Artificers, Surveyors, and Engineers.

PROP. 13-THEOR.

The angles which one st. line makes with another upon one side of it are either right angles, or together equal to two right angles.

CONSTRUCTION.-P. 11. To draw a st. line at right angles to a given st. line from a given point in the same.

DEMONSTRATION.-Def. 10. When a line standing on another line makes
the adjacent angles equal, each of the angles is a right angle.
Ax. 8. Magnitudes which exactly fill the same space are equal.
Ax. 2. If equals be added to equals, the sums will be equal.

Ax. 1. Magnitudes which are equal to the same, are equal to each
other.

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CASE II.-But suppose that the angle CBA is not equal to angle

ABD;

CONS. 1 P. 11.

2 Def. 10.

At B in DC draw BE at rt. angles to CD, thens CBE, EBD are two rt. angles.

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B

DBA =

D

E

F

G

/s DBE & EBA, to both add ABC;

. Ls DBA + ABC

=LS DBE, EBA

and ABC.

But s CBE, EBD D

==

/ s DBE, EBA

and ABC;

=

C

B

Ls CBE, EBD s DBA, ABC;
Now s CBE, EBD are two rt. angles,
Ls DBA, ABC together two rt. angles.
Wherefore, the angles which one line, &c.

Q.E.D.

SCHOLIUM.-1. A rt. angle FBG is formed by bisecting the angles ABD,

ABC.

2. If one angle be a rt. angle, the other is a rt. angle; if one be obtuse, the other is acute and if one be acute, the other is obtuse.

3. A semicircle is the measure of two right angles; and all the angles formed by any number of lines converging to one point, on one side of another line, are together equal to two right angles.

4. The supplement to an angle is what it is deficient of two rt. angles; thus, ABD is the supplement of angle ABC: the complement, what is wanting to make up one rt. angle; as, ABE is the complement of ang. ABC.

USE AND APPLICATION. This theorem is of frequent use in Trigonometry and Astronomy. When we know one of the angles which a st. line meeting another st. line makes, at a point in it, we in fact know the other, for the two angles are always equal to 180°; and if we subtract the given arc from 180° we have the other angle: thus, let ang. ABC equal 70°, ABD equals 180-70, or 110.

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