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IN

PROPOSITION XXXI. THEOREM XXVII.

N a circle, the angle (ADB) in a semicircle (ADEB), is a right angle; but the angle (DAB) in a segment (DAB) greater than a semicircle, is less than a right angle, & the angle (DEB) in a segment (DEB) less than a semicircle, is greater than a right angle: also the mixtilineal angle (BDA) of the greater segment, is greater than a right angle, & that (BDE) of the lesser segment, is less than a right angle.

CASE I.
Hypothesis.

Thesis
The V ADB is in the semi O ADEB.

This Y ADB is a L.
Preparation.
1. Draw the ray CD.
2. And produce AD to N.

Pof. 2. DEMONSTRATION,

1

Pos. 1.

BE CAUS

ECAUSE in the A ADC the side CA is = to the side CD (D. 15. B. 1.). 1. The V CDA is = to the V CAD.

P. 5. B. 1. Again, in the ACDB; the fide CD being = to the side CB, D.15. B. 1. 2. The V CDB is = to the V CBD.

P. 5. B. 1. 3. Consequently, V ADB is = to V CAD + CBD.

Ax. 2. B. i. But VNDB is also = to V CAD + CBD (P. 32. B. 1.). 4. Wherefore, this V NDB is = to V ADB.

Ax. 1.B.1. $. From whence it follows, that V ADB is a L.

D. 10. B. I.
CASE II.
Hypothesis.

Thesis.
The V DAB is in the segment DAB > a semi O.

This DAB is <aL DEMONSTRATION.

BECAUSE

E CAUSE in the A ADB, the V ADB is a L (Cafe I.). 1. The Y DAB will be <a La

R

P. 17. 2. 1.

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HE the opposite V DAB + DEB of the quadrilateral figure ADEB are = to 2 L.

P. 22. B. 3 2. Wherefore, V DAB being <a L (Case II.), DEB will be neceffarily > al

CASE IV.
Hypothesis.

Thesis.
The mixtilineal V BDA, BDE, are

The V BDA is > L,Elbe formed by the straight line BD E

V BDE is <a L. the arcbes DA, DE,

DEMONSTRATION.

ECAUSE the rectilineal V ADB, NDB, are L (Case I.). 1. The mixulineal V EDA will be necessarily > a L, & the inixtilineal V BDE <aL

Which was to be demonstrated

Ax. 8.B.I.

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PROPOSITION XXXII. THEOREM XXVIII. F a straight line (AB) touches a circle (ECF), & from the point of conta& (F) a chord (FD) be drawn; the angles (DFB, DFA,) made by this chord & the tangent, shall be equal to the angles (FED, FCD,) which are in the alternate segments (FED, FCD,) of the circle. Hypothefis.

Thesis. 1. BA is a tangent of the O ECF.

1. The FED is = to V DFB. II. And FD is a chord of this O

II. The V FCD is = to y DFA. drawn from the point of contact,

Preparation.
1. At the point of contact F in AB, erect the I fe.

P. 11. B.I. 2. Take any point C in the arch DF, & draw ED, DC, CF. Pol. 1.

DEMONSTRATION. ECAUSE the straight line AB touches the O ECF (Hyp. 1.), and FEis a Lerected at the point of contact F in the line AB (Prep. 1.). 1. The straight line FE is a diameter of the O ECF.

P. 19. B. 3. 2. Consequently, V FDE is a La.

P. 31. B. 3. 3, Wherefore, the V DEF + DIE are = to a La.

P. 32. B. i. But EFB or V DFE + V DFB being also = to a L (Prep. 1.). 4. The V DEF + DFE are = to the V DFB + DFE.

Ax. 1. B. i. Ś. Wherefore, the V DEF is = to V DFB, or the V in the segment s 4x. 3.B.1. DEF is = to the made by the tangent BF & the chord DF.

P.21. B. 3. Which was to be demonstrated. I. The V FED + FCD being = to 2 L (P. 22. B. 3.), & the adjacent V DFB + DFA being also = to 2 L (P. 13. B. 1.).

Ax.1. B 1. 6. The V FED + FCD are = to the V DFB + DFA. 7. Wherefore, V FED being = to the DFB ( Arg. 5.), the V FCD

is also = to the V DFA; or they in the segineni FCD is = to s Ax.3.B. 1. the contained by the tangent AF & the chord DF.

P.21. B. 3. Which was to be demonstrated. II.

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Pol. 3.

PROPOSITION XXXIII. PROBLEM V.

PON a given straight line (AB), to describe a segment of a circle (ADB) containing an argle equal to a given re&ilineal angle (N). Given

Sought The straight line AB together with V N. The segment ADB described uper

AB, containing an V = 10 V N. CASEI. If the given V is a L. (Fig. 1.). T fuffices to describe upon AB a semni O ADB. 1. This semi O will contain an V = to the given right V N. P. 31. B. 3. CASE II. If the given Vis acute (Fig. 2.) or obtufe (Fig. 3.)

Refolution. 1. At the point A in AB, make the VBAE = to the given VN. P. 23. B. 1, 2. At the point A in AE, erect the I AG.

P. 1. B. 1, 3. Divide AB into two equal parts in the point F.

P. 10. B. I. 4. At the point F in AB, erect the IFC, which will cut AG in C. P.U.B. 1. 5. From the center C at the distance CA, describe the O ADG. Pof. 3.

Preparation.
Draw the straight line CB.

Pos. 1. DEMONSTRATION. ECAUSE in the A ACF, BCF, the side AF is = to the side BF (ROf. 3.), FC common to the two A, & the V AFC = the V BFC (Ax. 10. B. 1. & Res. 4.). 1. The base CA is = to the base CB.

P. 4. B. r. 2. Consequently, the O described from the center C at the distance CA, will pass thro' the point B, & ADB is a segment described

D. 15.B.1. AB.

upon But AE touching the O ADB in A (Res. 2. & P. 16. Cor. B. 3.),

B. 1.

19. and AB being a chord drawn from this point of contact A ( Arg. 2.). 3. The V contained in the alternate segment ADB is = the V BAE. P. 32. B. 3.

Wherefore, V BAE being = to the given VN (Ref. 1.), the contained in the segment ADB described upor AB, is also = to the given V N.

Ax. 1, B. 1. Which was to be done.

BECAUSE

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PROPOSITION XXXIV. PROBLEM VI. To

O cut off a segment (BED) from a given circle (BDE), which shall contain an angle (DEB) equal to a given re&tilineal angle (N). Given

Sought The OBDE, & she rettilineal V N. The segment BED cut off from this ,

containing an V DEB=to ibegiven VN. Resolution.

P. 17. B. 3.

1. From any point A to the O BDE, draw the tangent ABC.
2. At the point of contact B in the line AB, make the V DBA

= to the given V N.

P. 23. B. I.

DEMONSTRATION.

Because the given V N is = to the V DBA (Ref. 2.), &

V DEB = to the V DBA (P. 32. B. 3.). 1. The V DEB & N are = to one another.

Ax. 1. B.. 2. Wherefore, the segment BED is cut off from the O BDE, containing an V DEB = io the given y N.

P. 21. B. 3: Which was to be done.

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