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PART. ENUN.-Let the three

of the ▲ ABC be bi

sected by the st. lines AD, BD, CD (Prop. IX.); produce AD to E (Post. 2), and from D draw DF to вC (Prop. XII.); then the ▲ BDE =

LCDF.

=

DEMONST.-1. Because the three Z of A ABC = two rt. 4 (Prop. XXXII.), .. the < DBA + DAB+ DCF = a rt. 4 (Const.) < DCF + CDF (Prop. XXXII.) ; by subtraction, 2 DBA + DAB = L CDF (Ax. 3).

2. But the exterior / BDE of ▲ ABD < DBA + DAB (Prop. XXXII.); B ... the BDE = Z CDF (AX. 1).

=

Wherefore, if the three angles &c.-Q. E. D.

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PROP. BB. THEOR.

GEN. ENUN.-If the sides of an equilateral and equiangular pentagon, or five-sided figure, be produced to meet, the angles formed by these lines are together equal to two right angles; and if the sides of an equilateral and equiangular hexagon, or six-sided figure, be produced to meet, the angles so formed are equal to four right angles.

PART. ENUN.-1. Let ABCDE be an equilateral and equiangular pentagon; produce

its sides to meet in F, G, H, K, L; then the 4 at these pts. together

rt. 2.

= two

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F

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rior

G and K (Prop. XXXII.) ;

H

BAF = / G, H, K, L; add to each the

AFB; .. the

at F, G, H, K," L = the three 4s of ABF = 2 rt. 4.

(Prop. XXXII.)

PART. ENUN.-2. Let ABCDEF be an equilateral and equiangular hexagon; produce its sides

to meet in G, H, K, L, M, N; the s

at these pts.

=

four rt. s.

DEMONST.-For they are the

8

4s of the two As GMK, HLN: and the 4 of any A

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two rt.

4s (Prop. XXXII.); .. the six Zs at G, H, K, L, M, N = four rt. s.

Wherefore, if the sides &c.

Q. E. D.

G

H

N

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PROP. XXXIII. THEOR.

GEN. ENUN. The straight lines, which join the extremities of two equal and parallel straight lines towards the same parts, are themselves also equal and parallel.

PART. ENUN.-Let AB, CD be = and || st. lines, and let them be A

joined towards the same

parts by the st. lines AC,

BD; then AC is = and || to BD.

CONST.-Join BC.

B

DEMONST.-1. Because AB = CD, and BC is common to the 2 AS ABC, BCD, ... the two sides AB, BC DC, CB, each to each and beto CD, the alternate BCD (Prop. XXIX.);

cause AB is alternate

AC

the base BD. (Prop. IV.)

2. Also the ACB = LCBD.

:

ABC =

the base

Hence, because the st. line BC meets the 2 st.

lines AC, BD, and makes the alternate ≤s ACB,
CBD = to one another, ... AC is || to BD.
And it has been shewn that AC = BD.
Therefore straight lines &c.-Q. E. D.

PROP. XXXIV. THEOR.

GEN. ENUN.-The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts.

N. B. A parallelogram is a four-sided figure, of which the opposite sides are parallel: and the diameter is the straight line joining two of its opposite angles.

PART. ENUN.—Let ABCD be a □m,

its diameter; then the opposite sides and s

are = to one another, and the diameter BC bisects it.

and BC

B

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D

DEMONST.-1. Because AB is || to CD, and BC meets them, .. the alternate ABC = alternate BCD (Prop. XXIX.): and because AC is to BD, and BC meets them, ... the alternate ACB = alternate CBD (Prop.

XXIX.); ... in the ▲ ABC, CBD, the 8 ABC, BCA = $ BCD, CBD, each to each, and the side BC, adjacent to theses, is common; ... the remaining sides are =, each to each, and the third to the third (Prop. XXVI.): i. e. AB = CD, AC = BD, and the BAC BDC. 2. And because the ABC = вCр, and

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ACB, ... the whole ABD = ACD: therefore the opposite sides

ands of ms are =.

3. Also because AB = CD, and BC is common to the ABC, BCD, .. the two AB, BC = DC, CB, each to each, and the included ▲ ABC = included BCD; .. the ABC = ▲ BCD (Prop. IV.): i. e. the diam. BC bisects the

m ABCD.

Wherefore the opposite sides &c.-Q. E. D. A few important Propositions are deducible from this Theorem, which, after first demonstrating its converse, so far as it is convertible, it may be proper to subjoin.

PROP. CC. THEOR.

GEN. ENUN.-If the opposite sides, or the opposite angles, of a quadrilateral figure be equal, the figure is a parallelogram.

PART. ENUN.-Let ABCD be a quadrilateral figure, of which the opposite sides, or opposite ▲, are = ; then ABCD is a m.

св.

DEMONST. First, let the opposite sides be

=. Join

1. Then, because AB = CD, and BC common to ▲ ABC,

DBC, .. the two AB, BC = DC,

CB, each to each, and the base
AC = base BD; .. the ABC

==

4 BCD (Prop. VIII.), and they are alternate ; .. AB is to CD. (Prop. XXVII.)

A

; .. also AC is to BD, and ...

2. Also the ACB CBD, and they are alternate ABCD is a m. (Def.)

Now the interior

3. Again, let the opposite / be =. Zs of the quadrilateral ABCD = 4 right 4 Cor. 1); also the 25 BAC + ACD =

(Prop. XXXII.

LS ABD + BDC

(Hyp.); .. the ▲ BAC + ACD = 2 rt. ≤ 3, and .'. AB is to CD. (Prop. XXVIII.)

4. In the like manner, it may be shewn that AC is to BD; .. ABCD is a □m.

Wherefore, if the opposite sides &c.-Q. E. D.

COR.-Hence it appears, from the definitions of the square, and the oblong, the rhombus, and the rhomboid, that these figures are ☐ms.

That the latter part of the 34th Proposition is not convertible, may be inferred from the following:

PROP. DD. THEOR.

GEN. ENUN.-A trapezium, of which the adjacent sides are equal, is bisected by the diameter.

PART. ENUN.-Let ABCD be a trapezium, of which the sides AB = BD, and CA = CD; then the diam. CB bisects it. DEMONST.-Because AB = BD,

and BC is common to As ABC,

DBC, .. the two AB, BC

= ᎠᏴ,

BC, each to each, and the base AC C

=

base CD; .. the ▲ ABC = A

DBC (Prop. VIII. Obs.); and the trapezium is bisected by diam. BC.

Wherefore a trapezium &c.-Q. E. D.

PROP. EE. THEOR.

A

E

GEN. ENUN.-The diameters of a parallelogram bisect each other; and, vice versa, if the diameters of a quadrilateral figure bisect each other, the figure is a parallelogram.

PART. ENUN.-Let ABCD be a parallelogram; its diams. AD, BC bisect each other in E, and vice versa.

DEMONST.-1. Because AC is

to BD, and CB meets them, ... the alternate / ACE = alternate EBD (Prop. XXIX.); and because AC is to BD, and AD meets them, ... the alternate ▲ CAE = alternate / EDB (Prop.

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