AB with the st. line CBE makes the /s ABC, ABE,
adjacent angles,

..theses ABC, ABE together two rt. angles:
But s ABC and ABD also = two rt. angles,
..s ABC and ABE = s ABC and ABD:
taking away the common ABC,

.. the rem. ABE the rem.

ABD,

and the less ABE the greater ABD; which is impossible;

BE is not in the same st. line with CB.

In the same way no line except BD is in the same st. line with CB;

.. BD is in one and the same st. line with CB. Wherefore if at a point in a line, &c.

Q.E.D.

SCH.-The words "upon the opposite sides of it" are of essential importance; for two lines may make with a third, two angles equal to two rt. angles, and yet the two lines not be in one st. line.

PROP. 15.-THEOR.

If two st. lines cut one another,the opposite, or vertical angles shall be equal.

5

Ax. 1.

Sub.

7

Ax. 3.

Sim.

& CEB = / AED.
...AE makes with CD the
8 CEA and AED,
..theses CEA & AED

together=twort.angles.

And. DE makes with AB the s AED, DEB;
..theses AED and DEB together

=

B

two rt. angles;

alsos CEA, AED = s AED, DEB:

Take away the common angle AED,

and rem. CEA will =rem. DEB.

In like manner we prove CEB = / AED.
Rec. Therefore, if two st. lines cut one another, &c.

Q.E.D.

COR. 1.-The angles formed by two st. lines, AB and CD, cutting each

other in one point E, are together equal to four right angles.

=

=

· Ls AEC and CEB together two rt. angles,
and s AED and DEB also : two rt. angles;
.. by adding the equals to the equals,
the s AEC, CEB, AED, and DEB together
angles.

=

four rt.

COR. 2.- And all the angles formed by any number of st. lines, AC, BC, DC, EC, FC, &c., diverging from a common centre, C, are together equal to four rt. angles,

alsos ACF, FCE, and ECG, togethers ACH and

HCG;

Con.

..all the angles diverging from C are together equal to four rt. angles.

SCH.-1. This proposition might be briefly proved by saying that the opposite ang les are equal, because they have the same supplement.

2. "This Prop. is the developement of the definition of an angle. If the lines at the angular point be produced, the produced lines have the same inclination to one another as the original lines have."-Potts' Euclid, p. 49.

3. The converse of this Prop. is, "If four lines meeting in a point make the vertical angles equal, each alternate pair of lines shall be in one and the same st. line."

APP.-To find the breadth of a lake, or the distance between two inaccessible objects, A and B.

From C, accessible both to A and B, measure the lines CA and CB, and produce them until CE is equal to CA, and CD to CB; join D and E, the distance DE will equal the distance AB.

For the lines CA, CB equal the lines CE, CD; and the angle ACB equals the angle DCE; therefore by Prop. 4, the lines AB and DE are equal.

A

D

PROP. 16.-THEOR.

E

If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

CON.-P. 10, P. 3, Pst. 1, Pst. 2.-DEM.-P. 15, P. 4, Ax. 9.

E. 1 Hyp.

2 Conc.

C. 1 by P. 10 & Pst. 1

2 Pst. 2, P. 3, Pst. 1.

D. 1 by C. 1 & 2,

234567

P. 15.
P 4.
C.
Ax. 9.

Sim. P. 10.
D. 1-5.

... ▲ AEB = A CEF, and Z BAE
But ECD or ACD > ECF;

and.. ACD > BAC.

G

FCE.

D

In like manner produce AC to G, and bisect BC, and BCG, i.e., ACD, will be > ABC. 8 Rec. Therefore if one side of a triangle be produced, &c. Q.E.D. PROP. 17.-THEOR. (converse of Ax. 12.)

Any two angles of a triangle are together less than two right angles.

E. 1 Hyp.

2 Conc.

Con.-Pst. 2, DEM.-P. 16, Ax. 4, P. 13.

Let ABC be a triangle ;

then any two s in it, as A
and B, are together <two rt.
angles.

C. 1 by Pst. 2. Produce BC to D.

D. 1 by Con..

ACD is the exterior angle,
ACDB or A.

Add. To each of these unequals add

the ACB

Ax. 4. then 8 ACD and ACB > /s ABC and ACB.
P. 13. But s ACD and ACB

D. 4... s ABC, ACB < two rt. angles.

Sim. In like manner s BAC, ACB < two rt. angles;
Sim. and s BAC, ABC <two rt. angles.

8

91 Rec.

Therefore, any two angles of a triangle, &c.

Q.E.D.

Scu. Both the 16th and the 17th propositions will be included in the thirty-second.

PROP. 18.-THEOR.

The greater side of every triangle is opposite to the greater angle.

6 Rec.

ADB > int. DCB,
ABD;

.. ABD > < DCB;

ABC > ACB.

Therefore, the greater side of every triangle, &c.

Q.E.D.

SCH.-The argument on which the conclusion depends is named "a fortiori," by the stronger reason, and proves that the given predicate belongs in a greater degree to one subject than to another; as in the Syllogism,-Y is greater than Z, and X greater than Y; much more is X greater than Z.

PROP. 19.-THEOR. (Converse of the 18th.)

The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

E. 1 Hyp.

2 Conc.

DEM.-P. 5. P. 18.

SUP.-The side AC must be greatcr than, equal to, or less than AB.

3

D. 1 & 2.

| If AC AB, then ABC = . ACB:

but

=

ABC / ACB;

ACAB.

4 Sup. & P. 18. Again, if AC is < AB, then ABC is< ACB:

6

D. 4 & 5.

Now, AC is neither equal to, nor less than, AB; .. AC is greater than AB.

9 Rec. Wherefore, the greater angle of every triangle, &c.

Q.E.D.

SCH.-1. Props. 5, 6, 18, and 19, may be combined into one thus,-" One angle of a triangle is greater than, equal to, or less than, another angle, as the side opposed to the one is greater than, equal to, or less than, the side opposed to the other; and vice versa."

2. By Props. 17, 18, and 19, we may prove that from the same point there can be drawn but one perpendicular to a given line, and that this perpendicular is the shortest of all the lines from the given point to the given line.

3. As from a given point only three lines can be drawn perpendicular to each other, it is impossible to imagine that there are more than three species of quantity,-a Line a Surface, and a Solid.

PROP. 20.-THEOR.

Any two sides of a triangle are together greater than the third side.

D. 2 C. 1 & 2 3 Ax. 9

4 Ax. 9.

5 P. 19.

6 C. 1.

7 D. 4 & 5. 8 Sim.

9

But BD = BA and AC together;

BA and AC together are > BC.

In like manner it may be proved that AB and BC are > AC,

and BC and AC > AB.

10 Conc. & R. Therefore any two sides of a triangle, &c.

Let AE bisect

Rec.

BAC.

BEA is > EAC, and CEA than
BA is > BE, and AC than CE:

Q.E.D.

EAB;

And BA+AC > BE + CE, which are equal to BC.
Therefore any two sides, &c.

Q.E.D.

COR.-The difference of any two sides of a triangle is less than the rem, side.

D. 1 by P. 20, 2 Sub.

3 Ax. 5.

The sides AC and BC are > AB;

take AC both from (AC + BC,) and also from AB;
.. BC is > AB diminished by AC, or than the difference
between AB and AC.

APP.-1. Of all lines that can be drawn from one point, A, to another E, and reflected to a third point, B, those are shortest, AE, EB, which make the angle of incidence, AEG, equal to the angle of reflection, BED.

C. 1 by P. 12 & Pst. 2. | From B draw BD

2 P. 3.

3 Assum., Pst. 1.

D. 1 by C. 1, 2.

2 P. 4. 3 Sim.

4 H. & C. & D. 2

5 C.

In triangles BED, CED, DE is common, BD equal to DC, and the angles at D equal,

BE is CE, and BED= / CED.

In the same way we prove BF to be = CF. Now s BED, DEC, and AEG are all equal; and AEC is a st. line:

6 D. 5 & C. 3, P. 20 Also AFC is a A; ... AF and FC > AC:

7 D. 2, 3.

8 Conc.

9 Rec.

but AC AE and EB, and FB = = FC;

.. AF, FC, or AF, FB, > AE and EB. Wherefore, of all lines that can be drawn &c. Q.E.D.

2. Natural causes act by the shortest lines; therefore all reflections are made by the lines which cause the angle of reflection to equal the angle of incidence. Hence, by means of a mirror placed horizontally, we may construct a triangle the perpendicular of which shall be representative of the height of any object.-See " Euclid Practically Applied," Part I., Pr. 20, I. p. 84.

C