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Pol. 2. Pol. 3.

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PROPOSITION II. PROBLEM II. ROM a given point (A), to draw a straight line (AL), equal to a given traight lire (BC). Given

Sought 1. The point A.

AL = BC. 2. The straight line BC.

Resolution. 1. From the point A to the point B, draw the straight line AB. Pof. 1. 2. Upon this itraight line AB conftru& the equilateral A ADB. P. 1.B.1. 3. Produce indefinitely the sides DA and DB of this A. 4. From the center B, at the diítance BC, describe O CGM. 5. And from the center D, at the distance DG, describe O GLN; Pof. 3. which cuts the straight line DA produced, somewhere in L.

DEMONSTRATION. ECAUSE the lines BC and BG, are drawn from the center B to the OGM (Ref. 4.). 1. Those two lines are rays of the fame O CGM,

D. 16. B. I. 2. Confequently, BC = BG.

D. 15. B. 1. And because the lines DG and DL, are drawn from the center D to

the O GLN (Ref. 5.). 3. Those lines, are also rays of the fame O GLN.

GLN.

D. 16. B. 1. 4. Consequently, DG = DL.

D. 15.B.I. But the lines DA 8 DB, being the sides of an equilateral

A ADB (Ref. 2.). 5. The line DA, is = to the line DB.

D.24. B. 1. Cutting off therefore from the equal lines DG, DL, ( Arg. 4.);

their equal parts DB, DA, (Arg. 5). 6. The remainder AL is = to the remainder BG.

Since therefore the line AL is = to the line BG (Arg. 6.), and the

line BC is also = to the same line BG (Arg. 2.). 7. The line AL is = to the line BC.

But it is manifeft that this line AL, is a line drawn from the given

point A (Ref. 3.). 8. Wherefore from the given point A, a straight line AL, equal to the given straight line BC, has been drawn,

Which was to be done,

Ax. 3.

Ax. I.

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PROPOSITION III. PROBLEM III.

WO unequal straight lines (A & CD) being given ; to cut off from the greater (CD) a part (CB) equal to the less A. Given

Sought the line CD > line A.

from CD to cut off CB = A.

Resolution.
1. From the point C draw the straight line CE = to the given
one A.

P. 2. B. 1. 2. From the center C and at the distance CE, describe O CEB ; Pof. 1. which cuts the greater CD in B.

DEMONSTRATION. He straight lines CB, CE, being drawn from the center C to the O BEF (Ref. 2.). 1. They are rays of the fame O BEF.

D. 16. B. 1. 2. Consequently, CB = CE.

D. 15.B.I. But the straight line A being = to the straight line CE (Ref. 1.);

and the straight line CB being likewife = to CE (Arg. 2.). 3. The straight line A is = to the straight line CB.

And since CB is a part of CD. 4. From CD the greater of two straight lines, a part CB has been cut off = to A the less.

Which was to be done.

The

Ax. i.

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PROPOSITION IV. THEOREM I. F two triangles (BAC, EDF,), have two sides of the one, equal to two sides of the other, (i, e. AB = DE, & AC = DF), & have likewise the angle contained (a) equal to the angle contained (d): they will also have the base (BC), equal to the base (EF); & the two other angles ( 6 & c) equal to the two other angles (e & f) each to each, viz. those to which the equal sides are opposite; and the whole triangle (BAC) will be equal to the whole triangle (EDF). Hypothesis.

Thesis. 1. AB = DE.

1. BC = EF. II. AC = DF.

II. V = Ve & Vo=Vf. III. Va= Vd.

III. A BAC = AEDF.

Preparation.
Suppose the ABAC to be laid upon the A EDF, in such a man-
ner that
1. The point A falls upon the point D.
2. And the side AB falls

upon

the side DE.

DEMONSTRATION. INCE the line AB is = to the line DE (Hyp. 1.), & the point A falls upon the point D (Prep. 1.), & the line AB upon the line DE (Prep. 2.). 1. The point B will fall necessarily upon the point E.

Because the Va= Vd (Hyp. 3.), & the point A falls upon the

point D (Prep. 1.), & the side AB upon the side DE (Prep. 2.). 2. The side AC will fall necessarily upon the side DF.

Moreover, since this side AC is = to the side DF. 3. The point C must fall also upon the point F. 4. Wherefore, the extremities B and C of the base BC, coincide with

the extremities E and F of the base EF. 5. And consequently, the whole base BC coincides with the whole base EF;

for if the base BC did not coincide with the base EF, though the extremities B and C of the base BC, coincide with the extremities E and F of the base EF ; two straight lines would inclose a space (EXF or EYF); which is impossible.

Ax. 12. Since therefore, the base BC coincides with the base EF (Arg. 5.).

Since

Ax. 9.

Ax. 9.

Ax. 9.

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6. This base BC will be = to the base EF.

Ax. 9.
Which was to be demonstrated. I
Again, the base BC coinciding with the base EF (Arg. 5.), & the two
other fides AB, AC, of A BAC, coinciding with the the two other

fides DE, DF, of AEDF (Prep. 2, Arg. 2.).
7. Those two ABAC, EDF, are necefsarily equal to one another.

Which was to be demonstrated. HI.
In fine, since the V6 & Ve to which the equal fides AC, DF are op-
pofite (Hyp. 2.); likewise, the Vc & f to which the equal fides
AB, DE, are opposite (Hyp. 1.), coincide both as to their vertices

and their fides (Arg. 1, 2, 3, 5.)
8. It follows, that the V6 & Ve, as also the Vc & Vf, to which the
equal fides are opposite, are equal to one another.

Which was to be demonstrated. II.

Ax 9.

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BECAUS

PROPOSITION V. THEOREM II. Nevery isosceles triangle (BAC): the angles (a & b) at the base (BC) are equal, & if the equal fides (AB, AC,) be produced: the angles lote & d +f) under the base (EC) will be also equal. Hypothesis.

Thesis. 1. The A BAC is an isosceles A.

1. Va & Vb are equal. II. AB & AC are produced indefinitely. II. Vete&Vd+f are also equal.

Preparation
1. In the side AB produced take any point D.
2. Make AE = AD.

P. 3. B. 1. 3. Through the points B & e, as alfo C & D, draw BE, CD. Pof. I.

DEMONSTRATION.
ECAUSE in the ADAC the two sides AD, AC, are equal to
the two sides AE, AB of EAB, each to each ( Prep. 2. Hyp. 1.);
and the V A contained by those equatsides is common to the two A.
1. The base DC is = to the base BE; & the two remaining Vm & b+d

of ADAC, are equal to the two remaining Vn&a+c of A EAB,
each to each of those to which the equal sides are opposite.
And because the whole line AD is = to the whole line AE (Prep. 2.),

and the part AB = to the part AC (Hyp. 1.); cutting off &c. 2. The remainder BD will be = to the remainder CE.

Again, since in the A DBC the sides DB, DC, are equal to the sides CE, EB, of A ECB, each to each (Arg. 2. and 1.), &

likewise V contained m is equal to V contained n ( Arg. 1.). 3. The two remaining of the one, are = to the two remaining Vof

the other, each to each, viz. Vote= Vd + f & Vd=tc.
The whole Va tc& b + d being therefore = to one another, as

also their parts Vc & Vd ( Arg. 1. & 3.); cutting off &c. 4. The remaining Va & b are likewise = to one another.

But those V are the two y at the base BC. 5. Therefore Va & Vb at the base BC are = to one another.

Which was to be demonstrated. I. Moreover, since Vetr=Vd+5 | Arg. 3.) are the v under the base. 6. It is evident that the Vetc & fd + funder the base, are also one another,

Which was to be demonstrated. II.

P. 4. B.1.

Ax. 3.

P. 4. B. 1.

Ax. 3:

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