PROP. XXXV. Plate I. Fig. 37.

Parallelograms, BCDA, BCFE, which land upon the same base BC, and between the same parallels AF, BC, are equal one to the other.

For AD (a) = BC (a) = EF, add DE common to both; (6) then is AE = DF. But alfo AB (4) = DC, and the angle A (c) = CDF. (d) Therefore is the triangle ABE=DCF. Take away DgE common to both triangles; (e) then is the Trapezium AB g D Eg CF; add B g C, common to both; (f) then is the parallelogram ABCD = EBCF. Which was to be demonftrated.

The demonstration of any other cafes, is not uniike, but much more plain and easy.

Schol. Fig. 7.

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If the fide AB, of a right-angled parallelogram ABCD, be conceived to be carried along perpendicularly thro' the whole line AC, or AC thro' the whole line AB, the area or content of the rectangle ABCD shall be produc'd by that motion. Hence a rectangle is faid to be made by the drawing or multiplication of two contiguous sides. For example; let AB be supposed four foot, and AC three; draw three into four, there will be produced twelve square feet for the area of the rectangle.

a 34. 1. b 2. ax.

C 29. 1. d 4.1.

e 3. ax.

f 2. ax.

This being supposed, the dimension of any parallelg- Fig. 37. gram (EBCF) is found out by this theorem. For the area thereof is produced from the altitude BA drawn into the base BC. For the area of the rectangle AC = parallelogram EBCF, is made by the drawing of BA into BC, therefore, &c.

PROP. XXXVI. Fig 37.

Parallelograms BCDA, GHFE, standing upon coual bafes BC, GH, and betwixt the same parallels AF, BH, are equal one to the other.

Draw BE and CF. Because BC (a) = GH (b) EF, (c) therefore is BCFE a parallelogram. whence

a hyp. b 34. 1.

GHFE.

C 33. 1.

d 35.8.

PROP.

the parallelogram BCDA (d) = BCFE (1) =

Which was to be demonstrated.

PROP. XXXVII. Plate I. Fig. 38.

Triangles, BCA, BCD, standing upon the same base BC, and between the same parallels BC, EF, are equal the one to the other.

(a) Draw BE parallel to CA, (a) and CF parallel to BD. Then is the triangle BCA (b) = half Pgr. BCAE (c) = half BDFC (6) = BCD. Which was to be demonstrated.

( PROP. XXXVIII. Fig. 38.

Triangles, BCA, DFC, fet upon equal bases BC, DF, and between the same parallels EF, BC, are equal the one to the other.

Draw EB parallel to AC, and DB parallel to FC. Then is the triangle BCA (a) = half Pgr. BCAE (6) = half BCDF (c) = CDF. Which was to be demonStrated.

Schol.

If the base BC be greater than DF, then is the triangle BAC DFC, and fo on the contrary.

PROP. XXXIX. Fig. 39.

1

Equal triangles BCA, BCD, standing on the same base BC, and on the same fide are also between the fame parallels AD, BC.

If you deny it, let another line AF be parallel to BC; and let CF be drawn. Then is the triangle CBF (a) = CBA (b)=CBD. (c) Which is abfurd.

PROP. XL. Fig. 40.

:

Equal triangles BCA, CFD, standing upon equal bases BC, CF, and on the same fide, are betwixt the same parallels.

If you deny it, let another line AH be parallel to BF, and let FH be drawn. Then is the triangle CFH (a) = BCA (6)=CFD. (c) Which is abfurd.

PROP. XLI. Fig. 38.

If a Pgr. AEBC have the fame base BC with the triangle BCE, and be between the fame parallels DE, BC, then is the Pgr. AEBC double to the triangle BCE.

Let

Let the line AC be drawn. Then is the triangle BCA (a) =BCE; therefore is the Pgr. ABCD (6) =2BCA (e) = 2BCE. Which was to be demonstrated.

Schol.

From hence may the area of any triangle BCE be found, for whereas the area of the Pgr. ABCD is produced by the altitude drawn into the base, therefore shall the area of a triangle be produced by half the altitude drawn into the base, or half the base drawn into the altitude; thus, if the base BC be 8, and the altitude 7, then is the area of the triangle BCE 28.

PROP. XLII. Plate I. Fig. 41.

To make a Pgr. ECGF equal to a triangle given ABC in an angle equal to a right-lined angle given.

Through A (a) draw AG parallel to BC, (b) make the angle BCG the angle given; (c) bisect the base BC in E, and draw EF parallel to CG; then is the problem refolved.

For (AE being drawn) the angle ECG is equal to the given angle by construction, and the triangle BAC (af) 2 AEC (c)=Pgr. ECGF. Which was to be done.

PROP. XLIII. Fig. 9.

In every Pgr. AGEL, the complements LD, GD, of those Pgrs. CK, KF, which stand about the diameter, are equal one to the other.

For the triangle AEL (a)=AEG, and the triangle ADC (a) =ADB; and the triangle DEK (a) =DEF (6) Therefore the Pgr. LDDG. Which was to be demonftrated.

PROP. XLIV. Fig. 42.

To a given right-line A, to apply a parallelogram FL, equal to a given triangle LBC, in a given angle C.

a 37. I.
b 34.1.
c 6. ax.

a 31. 1. b 23. 1.

CIO. 1.

d 38. 1. e 41. 1.

a 34. 1. b 3 ax.

(a) Make a Pgr. FD equal to the triangle LBC, so that the angle GFE may be equal to C. Produce GF till FH be equal to the line given A. Through H (6) draw IL parallel to EF, which let DE produced meet in I, let DG produced meet with a right-line drawn from I through F in the point K; thro' K (6) draw KL parallel to GH, which let EF continued meet at M, and IH at L. Then shall FL be the Pgr. required.

a 42. I.

b 31. I.

a

B 4

For

For the Pgr. FL (c) =FD LBC; (d) and the angle MFH_GFE=C. Which was to be done.

PROP. XLV. Plate I. Fig. 43, and 44.

Upon a right-line given FG, and in a given angle A, to make a Pgr. FL, equal to a right-lined figure given ABCD.

Resolve the right-lined figure given into two triangles BAD, ECD, then (a) make a Pgr. FH-BAD, so that the angle F may be equal to A. FI being produced, (a) make on HI the Pgr. IL BCD. Then is the Pgr. FL (6)=FH4-IL (c)=ABCD. Which was to be done.

Schol.

Hence is easily found the excess, whereby any rightlined figure exceeds a less right-lined figure.

PROP. XLVI. Fig. 6.

Upon a right-line given CD to describe a square DB. (a) Erect two perpendiculars CB, DA, (6) equal to the line given CD; then join BA, and the thing reqired is

done.

h29. dif.

f 33. 1. g cor. 29. 1. teral. Moreover the angles are all right, (g) because one A, is right; (b) therefore DB is a square. Which was to be done.

For, whereas the Angle C+D (c) = two right-angles (d) therefore are AD, BC parallel. But they are alfo (e) equal; (f) therefore AB, DC are both parallel and equal; therefore the figure AC is a Pgr. and equila

:

After the fame manner you may easily describe a rect

a 12. ax.

b 29. def.
€ 4. 1.
4.1. 1.

angle contained under two right-lines given.

PROP. XLVII. Fig. 45.

..

In right-angled triangles BAC, the Square BE, which is amade on the side BC that fubtends the right-angle BAC, is equal to both the squares BG, CH, which are made on the fides AB, AC, containing the right-angle.

Join AE, and AD; and draw AM parallel to CE. Because the angle DBC (a) =FBA, add the augle ABC common to them both; then is the angle ABD= FBC. Morcover, AB (6) FB, and BD (6) =BC; (c) therefore is the triangle ABD=FBC. But the Pgr. BM (d) 2 ABD,

2 ABD, and the Pgr. (d) BG
2FBC (for GAC is one
right-line, by Hypothesis, and 14. 1.) (e) therefore is the
Pgr, BM-BG. By the same way of argument is the
Pgr. CM=CH. Therefore is the whole BE= (f)BG+
CH. Which was to be demonstrated.

Schol.

This most excellent and useful theorem hath deferved the title of Pythagoras his theorem, because he was the inventor of it. By the help of which the addition and fubtraction of squares are performed; to which purpose serve the two following problems.

PROBLEM I. Plate I. Fig. 46.

To make one Square equal to any number of squares given. Let three squares be given, whereof the fides are AB, BC, CE. (a) Make the right-angle FBZ, having the sides infinite; and on them transfer AB and BC; join AC, then is ACq (b) = ABq + BCq. fer AC from B to X, and CE the third fide given from B to E; join EX. (6) Then is EXq = EBq (CEq)+BXq (ACq) (c) = CEq + ABq + BCq. Which was to be demonstrated.

ABq+BCq.

PROBLEM II. Fig. 2.

Then trans

Two unequal right-lines being given AE, EG, to make a Square equal to the difference of the two squares of the given lines AE, EG.

From the center E, at the distance of AE, describe a circle, and from the point Gerect a perpendicular FG, meeting the circumference in F; and draw EF. (a) Then is EFq (EAq) = EGq + FGq, (b) ThereEAq - EGq=GFq. Which was to be done.

Any two fides of a right-angled triangle ABC, being known, to find out the third.

Let the fides AB, AC, encompassing the right-angle, be, the one 6 foot, the other 8. Therefore, whereas ACq - ABq=64+36= 100 BCq, thence is BC √ 100

10.

Now