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D. 11 by P. 34. Because

2 Ax. 6. therefore

7s ABCD, DBCE are each double of a DBL,

ABCD = the DBCE.

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CASE II. Sup. Let AD and EF be not terminated in one point D. D. l¡byH.1&P.341 :: ABCD is a

TA DE FA E DF ... AD = BC; 2 H. 1, & P.34 and EBCF a o,

... EF = BC; 3 Ax. 1. and ... AD = EF. 4 Add.or Sub. Add, or take away the

common part DE; 5 Ax. 2 or 3. . , the whole, or rem.AE

B the whole, or rem.DF. 6 P. 34. And in As ÉAB, FDC, ::: AE = DF, and AB= DC, 7) P. 29. and the ext. _ FDC the int. and opp. _ EAB; 8 P. 4. ... EB = FC, and A EAB = A FDC. 9 Sub. From trapezium ABCF take away the equal as EAB

and FDC ; 10) Ax. 3. the rem. O ABCD = the rem. EBCF. 11 Rec. Therefore, Parallelograms upon the same base, &c. Q.E.D.

SCH.—1 The equality denoted is equality of surface, or area, not equality of sides and angles.

APP.-In the measurement of Surfaces or Areas, the unit of Surface is a rectangle; and it is therefore necessary to convert all parallelograms which are not rectangles into rectangles, in order to find their areas. The following method enables us to convert the parallelogram C

F ABDF into a rectangle ABCE, of equal area.

Produce indefinitely the parallel DF, and at 11 B and A, the extremities of the other parallel AB raise, by Prop. 11, the perpendiculars BE and AC; 3 then ABDF, will be converted into a rectangle ABCE, which by Prop. 35, is equivalent to ABDF. 5

Hence, the area of a parallelogram is equal to the area of a rectangle having the same base and altitude; and if we multiply the linear units A

B in the base by the linear units in the altitude, we obtain the square units, or units of surface in the parallelogram.

PROP. 36.—THEOR.—(A Corollary of the 35th). Parallelograms upon equal bases, and between the same parallels, are equal to one another.

H

Cox.-Pst. 1.-DEM.-P. 34, Ax. 1, P. 33, Def. A, P. 35 E. 1/ Hyp. 1. Let the Ds ABCD, A DE

EFGH, be on equal

bases, BC and FG,
2. and between the same is

AH and BG ;
2 Conc.
then o

ABCD

O EFGH.
by Pst. 1. Join the points B, E, and

C, H.
D

C.

EH ;

D. I byl 1&P34) BC=FG, and FG = 2 Ax. 1.

BC = EH. 3 H. 2 & C. But BC is || EH and BC and EH are joined towards the

same parts by st. lines BE, CH ; * P. 33. BE and CH are both equal and parallel ; 5 Def. A. and .:. also EBCH is a parallelogram. 6 D. 5, Now the Os EBCH, ABCD are on the same base BC

& H. 2. and between the same parallels BC, AH ;

P. 35. therefore O EBCH O ABCD. 8 P. 35. Also DEBCH = O) EFGH ; 9 Ax. 1. therefore o ABCD - EFGH. 10 Rec. Therefore, parallelograms upon equal bases, &c. Q.E.D.

APP.--1. The Diagonal Scale is constructed on the principle of parallelograms on equal Dases and between the same parallels being equal :

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The principle is thus shown,—“Let eb, kl, 1d, be three equi-distant parallel lines, having other equi-distant lines drawn at right angles across them. Join ad, then mg will be the half of cd or ab. For draw gh at rt. angles to cd, then the triangles amg, ghd are obviously equal, and hence mg hd=ch; that is, mg is the half of cd, or its equal ab.

“If, instead of drawing one line kl between eb, fd, there be drawn nine equidistant lines," as between the parallels KAB, GDC, “the part mg in the second line would obviously be 1-10th of ab, the part on the next line 2-10ths &c."—RITCAIE'S Geom., p. 29.

In the diagonal scale KBGC thus constructed, the distance AB represents 100, and each of the divisions between A and B 10; and on the diagonal line diverging from A, the first distance from the perpendicular AD to the diagonal will be 1-10th of 10, or 1; the secord distance from AD to the same diagonal, 2-10ths of 10, or 2 ; the third distance, 8-10ths of 10, or 3; and so on. Thus the spaces between E and A, are hundreds ; between A and B, tens; and between B and C, units. If, however, the spaces between E and A are tens, those between A and B are units, and between B and; C tenths; indeed the values depend on what we call the spaces between E and A.

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PROP. 37.-THEOR. Triangles upon the same base and between the same parallels are equal to one another.

CON.-Pst. 2, P.31, Def. A.--DEM.-P. 35, P. 34, Ax. 7. E. 11 Hyp. 1. | Let the As ABC, DBC be on the same base BC,

A D F 2. and between the same lls AD, BC; 2 Con. then the A ABC = the A DBC. c by Pst. 2 Produce AD both ways indefinitely; 2 P. 31. through B draw BE || CA,

and through C, CF || BD;.
3 Def. a. then EBCA and DBCF are parallelo-

grams.
D. I byC.3&H ::the Os are on the same baseBC,

and between the same || BC, EF;
2 P. 35. ... the OEBCA = the ODBCF :
3 P. 34.

is bisected by its diameter,
..A ABC = half the EBCA,

and A DBC = half the ODBCF ;
4 Ax. 7. ... also A ABC = A DBC.
51 Rec. Wherefore, triangles upon the same base, &c. QE.D.

SCH.-On the principle that triangles are the halves of parallelograms, the areas of triangles are obtained; the product of the base and altitude gives the area of a parallelogram ; consequently, half the product of the base and altitude gives the area of a triangle.

PROP, 38.-THEOR. Triangles upon equal bases and between the same parallels, are equal to one another.

Con.. Pst. 2, P. 31, Def. A.-Dem.-P. 36, P. 31, Ax. 7. E. 1| Hyp. 1. Let the As ABC, DEF, be

G A D

H н on equal bases BC, EF, 2. and between the same lls

BF, AD; 2 Conc. the A ABC = the A DEF. C. 1 by Pst. 2 Produce AD both ways in

definitely ; 2 P. 31. through B draw BG || CA,

B В K CE and through F,

FH | ED; 3 Def. A. then GBCA and DEFH are parallelograms. D. 1 by H.&Ci :: base BC = = base EF, and BF || GH ; 2 P. 36.

... the O GBCA the O DEFH : 3 P. 34. But a parallelogram is bisected by its diagonal ;

.. As ABC, DEF each = half GBCA, DEFH; 4 Ax. 7. and .. A ABC = A DEF. 5. Rec. Wherefore, triangles upon equal bases, &c. Q.E.D.

APP.-By the last two propositions we arrive at a practical way of dividing a triangular space, as ABC, into two equal parts; for if the base BC be bisected from the vertex A by AK, then, because the two triangles ABK and ACK are on equal bases BK and CK, and between the same parallels, AD and BC, those triangles are of equal areas.

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PROP. 39.-THEOR. Equal triangles upon the same base and upon the same side of it are between the same parallels.

Con.-Pst. 1, P. 31.-DEM.-P. 37, Ax. 1.
E. 11 Нур.
Let the ASABC, DBC,be equal,

A

D and on the same base BC,

and on the same side of it, 2 Conc. then the line joining A and D

shall be || to BC. C. 1 by Pst. 1. Join A and D, then AD is IMBC; 2 Sup.

but suppose them not to bë ll, 3 P. 31, Pst.1 through • A draw AE || BC, and join E, C.

B

с D. 1 byHyp.&C3: the As ABC, EBC are on the same base BC, and

between the same lls AE, BC; 2 P. 37.

the A ABC = the A EBC: 3 Hyp. But the A ABC also = the A DBC ; 4 Ax. 1.

the A DBC = the A EBC, 5 ex abs.

or, the greater A is equal to the less; 6 Conc.

AE is not parallel to BC. 7) Sim.

the same way it may be shown that no other line

except AD is par. to BC; 8 Conc. . AD is parallel to BC. 9 Rec. Therefore, equal triangles upon the same base, &c. Q.E.D.

SCH.-“If the vertices of all the equal triangles which can be described upon the same base, or upon equal bases as in Prop. 40, be joined, the line just formed will be a straight line, and is called the locus of the vertices of equal triangles upon the same base, or upon equal bases."

PROP. 40.-THEOR. Equal triangles upon equal bases in the same st. line, and towards the same parts, are between the same parallels.

CON.-Pst. 1, P. 31.-DEM.-P. 38, Ax. l.
E. 1 Hyp. 1. Let the equal A8 ABC,
DEF, be on equal bases

D
BC, EF ;
, 2. and let BC and EF be in

the same st. line, and to

wards the same parts ; 2 Conc. then AD is ll to BF. C. 1 by Pst. 1. Join A, D, and AD is

parallel to BF. 2) Sup. But suppose AD not parallel B CE

3P.31Pst.1 through A draw AG || BF, and join G, F; D. 1 by H.&C3: the As ABC, GEF are on equal bases, BC, EF, and

between the same lls, BF, AG;

M

to BF,

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D, 2] P. 38. ... the A ABC = the A GEF:

3 Hyp. 1. But A ABC also = A DEF;
4 Ax. 1. ..A DEF=A GEF,
51 er abs, or, the greater is equal to the less A, which is absurd :
6 Conc. AG is not parallel to BF.
7 Sim. In like manner no line except AD is || to BF ;
8 Conc, therefore AD is parallel to BF.
9| Rec.

Wherefore, equal triangles upon the same bases, &c. Q.E.D. APP:-1. A trapezium is equal to a parallelogram of the same altitude, and whose base is

B

C G half the sum of the parallel sides.

Let CD be bisected in H, and through H a parallel, GHF, to AB be drawn.

H Since CG and FD are parallel, the angles GCH and G are respectively equal to D, and HFD (29) and CH is equal to HD; therefore (26) CG is equal to FD, and the triangle

D CHG to the triangle DHF. Therefore AF and BG are together equal to AD and BC, and the parallelogram AG to the trapezium AC; and since AF and BG are equal, AF is half the sum of AD and BC

Thus the Area of a trapezium will be found, by taking half the sum of the parallel sides, and multiplying that half sum by the altitude CE.

2. The area of a square is found numerically by multiplying the number of equal parts in the side of the square by itself. Thus a square whose side is twelve inches, contains in its area 144 square inches. Hence in arithmetic, when a number is mulciplied by itself, the product is called its square. Thus, 9, 16, 25, &c., are the squares of 3, 4, 5, &c. ; and 3, 4, 5, &c., are called the square roots of the numbers, 9, 16, 25, &c. Thus square and square root are corelative terms."-LARDNER's Euclid, p. 54.

PROP. 41-THEOR.

If a parallelogram and a triangle be upon the same base and between the same parallels; the parallelogram shall be double of the triangle.

AE;

Con.-Pst. 1.-DEM.-P. 37, P, 84, E. 1 Hyp. 1 | Let the ABCD, and the A

D 4 EBC, both be on the

same base BC,

2 and between the same || BC, 2 Conc. then the ABCD shall be double of the A EBC.

B C. by Pst. 1. Join the points, A, C, by the

st. line AC. D. 1 byHyp1,2 . the As ABC, EBC are on the same base BC,

& between the same ||s, BC, AE,
2 P. 37. the A ABC = the A EBC :
8 P. 34. But s are bisected by their diagonals ;

Conc. ... the O ABCD = twice the A ABC;
D. 2. and O ABCD equals twice the triangle EBC.
Rec. Wherefore, if a parallelogram and a triangle, &c.

Q.E.D.

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