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Sch.—The converse of Prop. 41 is,--If a parallelogram is double of a triangle, and they have the same base, or equal bases upon the same straight line and towards the same parts; they shall be between the same parallels.
APP.-The general method for finding the area of a triangle, or of any figure that may be resolved into triangles, is founded on this proposition. The area of a parallelogram is measured by the product of the base and altitude ; and a triangle on the same, or on an equal base, being half the parallelogram, the area of a triangle is equal to the product of the altitude and half the base; or what is the same thing, equal to half the product of the base and altitude. The area of polygons is the sum of the areas of the triangles into which the polygons may be divided.
Dividing a circle by an infinite number of triangles having their common vertex in the centre, the area of a circle equals the product of the radius and of the semi-circumference.
To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
SOL.-P. 10, Pst. 1, P. 23, P. 31, Def. A.-DEM.-P. 38, P. 41, Ax. 6, E. 11 Dat. Given the A ABC and the angle D ; 2 Quæs. 1 to describe a O= the A ABC,
2 and having an angle = the angle. D. C. 1jbyP10,Pst 1| Bisect BC in E, and join AE ;
H A F G 2 P. 23. at E in EC make the <
- the ZD; 3 P. 31. and at A draw AFG ||
BC,& at C, CG || EF; 4 Def. A. then EFGC is a parallelogram,
D 5 Sol. and equal to the triangle ABC,
B В and having an angle =
angle D. D. I by C. 1 & 3. •• base BE : : base EC, and BC || AG ; 2 P. 38.
A AEC, and A ABC = twice A AEC : 3 C.
But the O FECG and the A AEC are both on EC,
and between the same ||s, EC, AG ; 4 P. 41. .. the parallelogram FECG = twice the A AEC ; 51 Ax. 6. and ... the O FECG the A ABC; 6 C. 2. | And Z CEF of D FECG = the < D. 7. Rec. Wherefore, there has been described a parallelogram, &c.
SCH.-To describe a trianglo equal to a given parallelogram,FECG and having an angle equal to a given angie D. As in the last figure, produce the parallels GF and CE, and make EB equal to EC ; ai B construct an angle equal to the given angle D, and produce BA to meet the
parallel GA in A, and join AC: then the triangle ABC thus constructed will, on drawing BH parallel to FE, be found by Prop. 41 equal to the parallelogram FEOG.
PROP. 43.-THEOR, The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another.
DEM.-P. 34, Ax. 2, Ax. 3.
and BK and KD the Ds, E
complements of the figure: 2 Conc. then the complement BK =
the complement KD. D. 1 by Hyp.1 ::: ABCD is a o, and AC its diameter ;
B 2] P. 34. ... the A ABC the A ADC: 3 Hyp. 2. And ::: AEKH is a O, and AK its diameter, 4 P. 32. ... the A AEK the A AHK : 5:P. 34. Also the A KGC= the A KFC. 6 D. 4 & 5. Then, :: A AEK: ΞΔ ΑΗΚ,
and A KGC=A KFC ; 7) Ax. 2. ... the two AS AEK and KGC=the two As AHK & KFC: 81 D. 2. But the whole A ABC = the whole A ADC ; 9) Ax. 3. therefore the rem. complement BK = the rem. complement
KD. 10 Rec.
Wherefore, the complements of the parallelograms, &c. Q.E.D. APP.-If any parallelogram, as KD in the last figure, be given, another parallelogram, KB, may be found, equal to it, and having one side EK, equal to a given line.
PROP. 44.--PROB. To a given line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given angle.
SOL.-P 42, P. 31, Pst. 1, Pst. 2.-DEM.-P. 29, Ax. 9, Ax. 12, P. 43, Ax. 1, P. 15 E. 11 Dat. Given the st. line AB,
A C, and <D;
the A C, and having с
=the < D.
the A C,
and having an ang. EBG = _D;
F. 2 Position. place the
so that the
line with AB;
|| BG or EF, and join
L 5 Pst. 2. Next produce HB and
FE to meet in the point K;
C. 6 P.31, Pst. 2 through K draw KL || AE or HF, and produce GB and
HA to M and L 71 Sol. then the O BL= the AC,
and has its ang. ABM = _D. D. 1 by C.3&4. .:: HF falls on the parallels AH and EF,
2 P. 29. the <s AHF and HFE together = two rt. angles ; 3 Ax. 9. and Zs BHF and HFE together are < two rt. angles ; 4 Ax. 12. but when int. angles on the same side are < two rt.
angles, their sides meet, if produced ; 5 Conc. .. HB and FE, being produced, will meet : 6 Pst. 2. Let them be produced and meet
in the point K; 7 P. 31. through K draw KL || EA or FH; 8 Pst. 2. and produce HA to L, and GB to M; 9 Conc. then FKLH is a O, HK the diameter, AG and ME
7s about HK, and LB and FB the complements ; 10 P. 43. and ... compl. LB = compl. BF: 11C.1 & Ax.1 But O BF=AC,..LB = A C. 12 P. 15 & C. 1 And :: _ GBE = L ABM, and < GBE = _D, 13 Ax. 1.
also Z ABM Therefore, to the st. line AB the parallelogram, &c. Q.E.D.
APP.-This proposition contains a kind of Geometrical Division.
Arithmetically we can say, a given triangle contains, for example, 20 square feet,-a given line 4 lineal feet; how can we apply a parallelogram to the 4 lineal feet, so that the area of a parallelogram shall equal that of the triangle? We divide 20 by 4, and the quotient 5 is the lineal measurement of the other side of the parallelogram.
Geometrically we say, a given rectangle BC contains 12 square feet,-a given line BD, or its equal EH, 2 lineal feet; how can we apply another rectangle to BD or EH measuring 2 lineal feet, so that the area of the required rectangle may equal the area of the given rectangle? Produce indefinitely the parallels AC and BE, A C
F AB and CE; from B set off BD equal to the representative
value of 2 feet, and draw the diag. DE; produce DE until it cuts AC produced in F; the
diagonal DF will so divide or intersect AC produced, that the part CF shall equal the other side of the rectangle, of which the given side BD is 2 lineal feet.
B For, complete the rectangle ADGF; EH equals BD, and EI equals CF; and the comple
E ment EG is equal to the complement EA: and if we divide AE into equal squares, and EG into D equal squares, we find that in the first there are
H 12, and in the second 2 X 6, or 12 also.
PROP. 45.- PROB. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.
Sol.-Pst. 1, P. 42, P. 44.- DEM.---Ax. 1, Ax. 2, P. 29, P. 14, P. 30, Def. A.
E, I Dat. Given a rectil fig. ABCD, and a rectil. < E;
A D F GL 2 Quæs. to describe a
= fig. ABCD, and having an angle
E C. 1 by Pst. To divide the fig. into AS,
join D,B; 2 P. 42. describe a O FH = the
KH MI A ADB, and having an
_ FHK = _E; 3 P. 44. to the st. line GH apply the GM = the A DBC, and
having an _ GHM = _E; 4) Sol. then the fig. FKML is the o required. D. 1 byC.2&3 ZE < FKH, and also Ż GHM ; 2 Ax. 1.
< FKH < GHM : 3 Add.
To each of the equal angles add the Z KHG ; 4 Ax. 2. then the Zs FKH + KHG = <s GHM + KHG ; 5 P. 29. But s FKH and KHG = two rt. angles; 61 Ax. 1.
<s GHM and KHG = two rt. angles : 7 D. 6. Thus at the · H in the st. line HG, the adjacent <s GHM
= two rt. angles ; 8 P. 14. .. KH is in the same st. line with HM.
9 C. Again, •:• HG meets the parallels KM and FG, 101 P. 29.
i. _ MHG < HGF; 11 Add.
To each of the equal angles add the _ HGL. 12) Ax. 2. then _s MHG #HGL= Ls HGF + HGL : 13 P. 29. But s MHG + HGL = = two rt, angles ; 14 Ax. 1. ZS HGF + HGL = two rt. angles, 151 P. 14. and ... FG is in the same st. line with GL. 16 C. 2 & 3 And :: KF || HG, and HG || ML; 17 P. 30. .-. KF is parallel to ML: 18 C. 2 & 3 Also KM is parallel to FL ; 19 Def. A. ::. fig. KFLM is a parallelogram. 20 C. 2 & 3 And: A ABD=D HF, and A BDC=O GM ; 21 Ax. 2. therefore the whole rectil. fig. ABCD = the whole
KFLM. 22 Rec. Wherefore, the parallelogram KFLM has been described, &c.
Q.E.F. COR.--Hence a parallelogram equal to a given rectil. fig. can be applied to a given right line and in a given angle, by applying to the given st. line a parallelogram equal to the first triangle. ABD p. 44, and having an angle = the given angle.
APP.--1. By this and the preceding Problem we may measure the superficial content of any rectilineal figure whatever, by first reducing it to triangles, and then making a rt. angled parallelogram equal to the sum of the triangles. We may also make a rt. angled parallelogram on a given line, and which shall be equal in area to several irregular figures. Also if we have several figures, we may make another equal to their difference.
2. By principles already established, and by the problems for the conversion of rectilineal figures into parallelograms of equal area, we may change any right-lined figure, as ABCDE, first into a triangle, and then into a rectangle of equal area.
Join DA and DB to divide the given figure into triangles, and produce AB indefinitely. Through E draw EH parallel to DA, and through C, CF parallel to DB; join DH and DF; then the triangle DHF is equal in area to the given figure ABCDE.
E Next, through D draw DN parallel to HF; bisect HF, a side of the triangle DHF, in L; at L raise the perpendicular LM, and through F draw FN a parallel to LM; the fig. LMNF is a rectangle of the same altitude as the triangle, and on half its base, and is therefore equal in area to the triangle DHF.
To describe a square on a given st. line.
SOL.--P. 11, P. 3, P. 31, Def. A.--DEM.--P. 34, Ax. 1, P. 29, Ax. 3, Def. 30. E. 1 Dat.. Given the st. line AB ;
2 Quæs. on AB to draw a square.
2 P. 3. make AD = AB;
and through · B, BE || AD;
A 1). 1 by C. 4. fig. ABDE is a parallelogram,
2 P. 34 & C.2 AB = DE, AD = BE, and AB = AD; 3 Ax. 1.
AB = AD = DE = EB ; 4 D. 1 & 2. and .. the OABDE is equilateral. 5 C. 3. Also, • : AD meets the parallels AB and DE, 6 P. 29. ... the int. _8 BAD + ADE = two right angles : 7 C. 1. but the _ BAD is a rt. angle. 8 Ax. 3.
LADE is also a rt. angle. 9 P. 34. Now the opposite angles of Os are equal ; 101 D. 7 & 6. ii. _ ABE opposite to < ADE is a rt. angle,
and Z BED opposite to Z BAD is a rt. angle; 11 D. 6, 7, & 9. .. the figure ADEB is a rectangle ; 12 D. 3. it is also equilateral ; 13 Def. 30. therefore 0 ABED is a square on AB.
Cor. 1.—The squares on equal lines are equal; and if the squares are equal, the lines are equal.
2. Every parallelogram having one rt, angle, has all its angles rt, angles.
APP --The Geometrical Square is an instrument by moans of which, and of the property of similar triangles that the sides about the equal angles are proportional, the height of an inaccessible object can be ascertained, provided a measurement to the perpendicular from The object can be made.-- See Euclid Practically Applied, Prop, 46, book 1.