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Part. EnUN.—Let the three /s of the A ABC be bisected by the st. lines AD, BD, CD (Prop. IX.); produce Ad to E (Post. 2), and from draw df I to BC (Prop. XII.); then the LBDE = L CDF.

DEMONST.-1. Because the three /s of A ABC = two rt. 28 (Prop. XXXII.), .. the ZS DBA + DAB + DCF = a rt. 2 (Const.) 28 DCF + CDF (Prop. XXXII.) ;

subtraction, ZS DBA + DAB = L CDF (Ax. 3). 2. But the exterior / BDE of A ABD

28 DBA + DAB (Prop. XXXII.); B .. the / BDE = L CDF (Ax. 1).

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

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PROP. BB. THEOR.

F

А.

GEN. Enun.-If the sides of an equilateral and equian. gular 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 equi. angular pentagon; produce its sides to meet in F, G, H, K, L; then the 4s at these pts. together two rt. ZS.

DEMONST.–For the L ABF (the exterior L of A LBH) 18 at h and L,

K and the L BAF (the exterior L of A GAK)

Sat G and K (Prop. XXXII.) ; .., by addition, ZS ABF + BAF = SG, H, K, L; add to each the SAFB; ... the 2 at F, G, H, K," L the three /s of ABF 2 rt. 1$. (Prop. XXXII.)

B

E

D

H

N

G

F

M

Part. ENUN.—2. Let ABCDEF be an equilateral and equi. angular hexagon; produce its sides to wees in G, H, K, L, M, N; the 28 at these pts.

four rt. ZS. Demonst. For they are the <s of the two AS GMK, HLN: and the /s of any A two rt. {* (Prop. XXXII.); .. the six L' at G, H, K,

M, N =

four rt. Ls.

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

E

н

D

K.

PROP. XXXIII. THEOR.

A

B

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 joined towards the same parts by the st. lines AC, BD; then AC is

and ll to BD.

Const.-Join Bc.

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 because AB is || to cd, the alternate < ABC = alternate BCD (Prop. XXIX.); .. the base AC = the base Bd. (Prop. IV.)

2. Also the 2 ACB = Z CBD. Hence, because the st. line Bc meets the 2 st.

lines AC, BD, and makes the alternate ZS 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.

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А.

B в

are

D

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 om,

and BC its diameter; then the opposite sides and 28

= to one another, and the diameter Bc bisects it.

DEMONST.–1. Because AB is || to cd, and BC meets them, .. the alternate < ABC = alternate 2 BCD (Prop. XXIX.): and because AC is || to bd, and Bc meets them, ... the alternate < ACB = alternate z CBD (Prop. XXIX.); .. in the As ABC, CBD, the < $ ABC, BCA = Z $ BCD, CBD, each to each, and the side BC, adjacent to these = < $, is common; : . the remaining sides are =, each to each, and the third < to the third 2 (Prop. XXVI.): i. e. AB = CD, AC = BD, and the Z BAC = 2 BDC.

2. And because the < ABC = < Bcp, and

the Z CBD = L ACB, ... the whole 2 ABD = the whole 2 ACD: therefore the opposite sides and Zs of oms are =.

3. Also because AB = CD, and Bc is common to the As ABC, BCD, ... the two AB, BC = DC, CB, each to each, and the included 2 ABC = included 2 BCD; .. the A ABC = (Prop. IV.): i.e. the diam. BC bisects the om 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.

Δ BCD

PROP. CC. THEOR.

А

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 2$, are = ; then ABCD is a om

DEMONST.–First, let the opposite sides be =. Join св.

1. Then, because AB = CD, and Bc common to AS ABC, DBC, the two AB, BC = DC,

B CB, each to each, and the base

base BD; .. the L ABC

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

2. Also the L ACB = L CBD, and they are alternate <8; .. also ac is ll to BD, and .. ABCD is a om. (Def.)

3. Again, let the opposite 25 be =. Now the Z interior 28 of the quadrilateral ABCD = 4 right 48 (Prop. XXXII. Cor. 1); also the ZS BAC

LS ABD + BDC

AC =

D

+

ACD

(Hyp.) ; .. the 28 BAC + ACD = 2 rt. 28, and ... AB is Il to cd. (Prop. XXVIII.)

4. In the like manner, it may be shewn that ac is || to BD; . . ABCD is a om.

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 ams.

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

PROP. DD. THEOR.

A

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 A8 ABC, DBC, .. the two AB, BC = DB, BC, each to each, and the base AC C

base CD; ... the A ABC = DBC (Prop. VIII. Obs.); and the trapezium is bisected by diam. Bc.

Wherefore a trapezium &c.—Q. E. D.

E

PROP. EE. THEOR.

Gen. Enun.—The diameters of a parallelogram bisect each other; and, vice versá, if the diameters of a quadrilateral figure bisect each other, the figure is a parallelogram.

Part. Enun.—Let ABCD be a parall am; its diams. AD, BC bisect each other in E, and vice versa.

DEMONST.-1. Because Ac is Il 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 Z CAE=alternate / EDB (Prop.

A

B

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