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PROP. 5.-THEOR.

The angles at the base of an isosceles triangle are equal to each other; and if the equal sides be produced, the angles on the other side of the base shall be equal.

CON.-Pst. 2, P. 3, Pst. 1.-DEM.-P. 4, Ax. 3, Def. 24, Ax. 1.

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E. 1 Hyp. 1.

F

Let A B C be an isosc. A of which side AB = side AC; 2 H. 2 & Pst. 2. and let the equal sides be produced indefinitely to D

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and E;

then the s ABC, ACB, at the base, are equal;
and the

s DBC, ECB, on the other side of the base, are also equal.

In AD take any F, and make AG = AF;
join the s F and C, G and B.

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=

D. 1by C.1& Hypl . AF AG, AC

=

AB, and A is common to

the two As AFC, AGB;

=

.. the side FC GB, and ▲ AFCA AGB; also ACF = ABG, and AFC = / AGB. C.1&Hyp.1 Again, the whole AF the whole AG,

2

P. 4.

3

D. 2.

4

5

Sub. Ax.3.

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and the part AB = the part AC,
on taking away the equals AB and AC,
the rem. BF =
= the rem. CG.

But in As BCF, BCG, BF: = CG, and FC = GB,
and BFC or AFC = CGB or AGB;
.. A BFC = A CGB, FBC = LGCB,
and BCF = / CBG.

D. 1by Superp. | Applying ▲ ABC to A DEF, so that A is on. D,

2 D.1, H.1

3 Ax. 8.

4 D.2,3,H.2 5 Conc.

6 H.1,&Conc. 7 D. 3, 6.

8 Con. sup.

9 Ax. 10.

10 Ax. 8.

and AB on DE;

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Again,. AB coincides with DE, and ▲ BAC= / EDF,

... the line AC shall fall on the line DF:

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But AC being DF, ... the C shall fall on the F; and B falling on E, and C on F, the line BC falls on the line E F:

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For if, though B falls on E, and C on F,BCdoes not fall on EF, then two st. lines will enclose a space;

But this is impossible;

.. the base BC does coincide with and the base EF: 11 D.2. 5, 6, 10 And A B falling on and being equal to D E; AC to DF;

12 Ax. 8.

13 Hyp. 1, D.10

and BC to EF,

A ABC coincides with and

A DEF.
Also, since D E coincides with AB, and EF with BC,
the ABC shall coincide with and equal / DEF.
And in a similar way ACB=/ DFE.
Wherefore, if two triangles have two sides, &c. Q.E.D.

14 Ax.8, Conc. 15 Hyp.1, D. 7. 16' Rec.

SCH.-1, This being the first Theorem, it is exclusively proved by means of the Axioms. The converse of the 8th Axiom is assumed; namely, that if the magnitudes are equal, not merely if they are equivalent, they will coincide.

2. The equality spoken of is equality of the sides and of the angles. Triangles may be equal in area, though the sides and angles of the one are not equal to the sides and angles of the other. When the sides and angles mutually coincide, the triangles are named equal triangles; when their areas only are equal, equivalent triangles.

APP.-This proposition contains the first of the criteria by which to infer the equality of triangles, and is applied to various uses: as

1st. In all parts of Geometry to establish the equality of triangles.

2nd. To ascertain an inaccessible distance, as A B, the breadth of a lake.

Take the angle at C formed by the lines from the extremities, A and B, of the inaccessible distance; and measure unn the distances CA and CB. The representative Values of these measure- A ments must now be taken from a Scale of Equal Parts, and drawn on paper, or on any plane surface: thus, draw a st. line DF of an indefinite length, and at D, by aid of the graduated semicircle, form an angle FDE, equal to the angle BCA: from a scale of equal parts represent the distance from A to C, in proper proportion, by the line

C

7B E

DE, and the distance from B to C, by DF; consequently, on a principle established in the Sixth Book, Prop. 4, and which we now assume as a Lemma,-" that the sides about similar triangles are proportional,"-the line BF will represent the distance from A to B and if to the same scale we apply the line EF, that distance on the scale will be the representative easurement of the actual distance AB,

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PROP. 5.-THEOR.

The angles at the base of an isosceles triangle are equal to each other; and if the equal sides be produced, the angles on the other side of the base shall be equal.

CON.-Pst. 2, P. 3, Pst. 1.-DEM.-P. 4, Ax. 3, Def. 24, Ax. 1.

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E. 1 Hyp. 1.

Let A B C be an isosc. ▲ of which side AB: = side AC;

2 H. 2 & Pst. 2. and let the equal sides be produced indefinitely to D

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C. 1 by P. 3.

2 Pst. 1.

and E;

then the

s ABC, ACB, at the base, are equal;

and the s DBC, ECB, on the other side of the base, are also equal.

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In AD take any F, and make AG = AF;
join the s F and C, G and B.

D. 1by C.1& Hypl. AF =AG, AC= AB, and A is common to

D. 2.

the two As AFC, AGB;

.. the side FC GB, and A AFC:
=

also ACF= =

A AGB;
ABG, and ▲ AFC = ▲ AGB.
C.1&Hyp.1 Again, . the whole AF = the whole AG,
and the part AB = the part AC,

2 P. 4.

3

4

5

Sub. Ax.3.

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on taking away the equals AB and AC,
the rem. BF: the rem. CG.

But in As BCF, BCG, BF = CG, and FC = GB,
and BFC or AFC = CGB or AGB;
BFC = ▲ CGB, FBC =
and BCF = / CBG.

.. A

Z GCP

9, D. 3 & 8.

ABG

the whole / ACF, and

Now the whole
the part CBG the part BCF;

10 Sub.& Ax. 3... on taking away Zs CBG and BCF,

11 Remk.

12 D. 8. 13 Remk.

the rem. ABC = the rem.
and these are the angles at the base.

ACB,

GCB,

And FBC was proved to be equal to

and these are the angles below, or on the other side of the base.

Wherefore, the angles at the base, &c.

COR.-Every equilateral triangle is also equiangular.

14 Rec.

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Let ABC be an equil. A,

A B B C A C;

then shall its angles be equal,

LA to B to /C.

Since A B-A C,

Q.E.D.

.. <B=/C,

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and since CA=CB,

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Ax. 1.

consequently, C= ▲ A.

Conc.

thes are all equal, A to B, B to C, and C to A.

Rec. Wherefore, if a ▲ be equilateral, &c.

(PROP. 6.-THEOR. (Converse of P. 5.)

Q.E.D.

If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to the equal angles, shall be equal to one another. CON.-P. 3, Pst. 1.-DEM.-P. 4,Ax. 1.

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DC = AB, and A DBC = ▲ ABC.
Thus the less is declared: the greater, which is absurd;
AB is not AC, that is, AB = AC.
Wherefore if two angles of a triangle, &c.

COR.-Every equiangular triangle shall also be equilateral.

Q.E.D.

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LBC, ... AC AB.

and '.' / B= ▲ A,... AC=BC;

and .. AB = BC.

Hence the side AB = BC, BC CA, and CA = AB. Wherefore, every equiangular triangle, &c.

Q.E.D.

SCH. 1. Converse theorems are not universally true; for instance, the direct proposition is universally true, "If two triangles have their three sides respectively equal, the three angles of each shall be respectively equal;" but the converse is not universally true,-" If two triangles have the three angles in each respectively equal, the three sides are respectively equal."-Potts' Euclid, p. 48. For equality in triangles one side at least of the one triangle must be given equal to one side of the other triangle.

2. There are two modes of Demonstration,-the direct, showing why a thing is so; and the indirect, proving that it must be so. Direct Demonstration, as in Prop. 5, is that in which we find intermediate steps, which proceed regularly to prove the truth of the proposition: the Indirect Method is only employed, as in Prop. 6, when the predicate of it admits of an alternative, and one of them must be true, because they exhaust every case that can possibly exist. We prove that the alternative cannot be true, and infer therefore the predicate must be true.

APP.-Hieronymus, records that THALES, who was living 536 B.C., measured the height of the pyramids of Egypt, by observing the shadows which they cast.

The height of an object, AB, and the length of its shadow BC are the same, when the light, which the object, AB, intercepts, is at an elevation of 45°: for the angles BCA, BAC, being each half a rt. angle, the sides which subtend them are equal. Thus by measuring the shadow CB, we obtain the height BA.

The height would also be obtained by making an observation with a quadrant of altitude.

PROP. 7.-THEOR.

Upon the same base and upon the same side of it there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those equal which are terminated in the other extremity.

CON.-Pst. 1, Pst. 2.-DEM-P. 5, Ax. 9.

E. 1 Hyp. 1. On AB, and on the same side of it, let

2

2.

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3 Conc.

there be two As, ACB, ADB;
and let the side CA = the side DA:
then it is impossible that CB should
equal DB.

SUP. If possible, in As ACB, ADB, let CA = DA,

and CB: = DB.

A

CASE I.-Let the vertices C and D be without each other.

C. 1by Pst. 1. Join C and D.

D. 1byHyp.&P.5

2

C.&Ax. 9.

AC AD, ../ ACD = / ADC:
But / ACD > BCD,

ADC > BCD:

à fort.
H. & P. 5.
5 D. 3.

BCD;

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D. 3, 4.

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D

BDC is both >and=_BCD, which is impossible

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