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Pol. 2. Pol. 3.
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.
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,
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.
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.
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.
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.).
6. This base BC will be = to the base EF.
fides DE, DF, of AEDF (Prep. 2, Arg. 2.).
Which was to be demonstrated. HI.
and their fides (Arg. 1, 2, 3, 5.)
Which was to be demonstrated. II.
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.
P. 3. B. 1. 3. Through the points B & e, as alfo C & D, draw BE, CD. Pof. I.
of ADAC, are equal to the two remaining Vn&a+c of A EAB,
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.
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.
P. 4. B. 1.