will be one right angle; that is, the angles F AC, A F B, B F C will be one right angle (for A F C is equal to A F B, BF C). In like manner it is proved, that the angles CF D, C D F are equal to one right angle. Therefore the angles FAC, AFB, BFC are equal to the angles CF D, CDF. And if the angle B F c be added in common, the angles FA C, A F B, and twice the angle B FC, will be equal to the angles BFC, CFD, CDF; that is, to the angles BFD, CDF. And if the equal angles A F B, B F D be taken away, there will remain the angle FA C, and twice the angle B F C, equal to the angle C D F. Therefore the angle C D F exceeds the angle F A C by twice the angle B FC; and fo the angle BFC is one half their difference.

Secondly, Let the perpendicular F c fall without the triangle AF D. Then because the external angle A D F of the triangle A F D is [by 32. 1.] equal to the internal angles DFC, DCF, and the right angle D C F is equal to the angles FAC, AFC; the angle ADF will be equal to the angles FAC, AFC, DFC; that is, equal to the angles FAC, AFD, and twice the angle D F C; or to the angles F A C, twice. the angle B F D, and twice the angle DFC; or equal to the angle FA C, and twice the angle B F C. Therefore the angle A D F exceeds the angle FA C, by twice the angle B FC; and fo the angle BFC will be one half the difference of them.

Wherefore in every triangle, the angle contained under the perpendicular drawn from the angle oppofite to the bafe upon it, and the right line bifecting that angle, will be one half the difference of the angles at the base. Which was to be demonstrated.

SCHOLIU M.

When the perpendicular F C falls without the triangle, it is evident, that the vertical angle AFD is the difference of the angles AFC, D F C at the perpendicular. But when the perpendicular F C falls within, the angle B F C itfelf (which is one half the difference of the angles at the bafe) is also one half the difference of the angles CF D, A F D at the perpendicular. For because the angles AFB, BFD are [by Luppofition] equal; if the angle BFC be added to both, the angle AFC will be equal to the angles BFD, BF C; that is, to the angle CDF, and twice B FC: wherefore the angle B F C is one half the difference of the angles AFC, CF D. PROP.

PROP. VII. THEOR.

If there be any right-lined figure having fix fides, and two contiguous fides of it be alternately equal and parallel to two other contiguous and oppofite fides, each to each; I say, first, the two remaining fides will be equal and parallel, fecondly, the oppofite angles (viz. the first and fourth, fecond and fifth, third and fixth) will be equal to one another; thirdly, any diagonal joining two of thofe oppofite angles, will divide the figure into two equal parts.

G

B

C

For let ABCDEF be a right-lined figure having fix fides; and let the two contiguous fides A B, BC be alternately equal and parallel to the two oppofite contiguous fides FE, ED; that is, D F equal and parallel to B C, and E D equal and parallel to A B: I fay, firft, that the remaining fides CD, AF will also be equal and parallel. Secondly, the oppofite angles ́A, D; B, E; C, F; will be respectively equal to one another, Thirdly, the diagonal B E will

H

F

D

E

divide the figure into two equal parts.

For draw the diagonals A E, AC, BF, BD, CE, DF; and let the diagonal BE cut the diagonals A C, D in the points G, H.

Then because the fides A B, E D are equal and parallel [by 33. 1.] the figure ABDE will be a parallelogram, having its fides A E, B D equal and parallel, and the angles A BE, BE D equal. In like manner, because the fides B C, EF are equal, the figure FBCE will be a parallelogram, having the oppofite fides FB, E c equal and parallel, and the angles FE B, C B E equal. Therefore the angles A B E, E B C will be equal to the angles B E D, FEB; that is, the whole angle ABC equal to the whole angle FED. Where

fore

fore fince there are two triangles A B C, DE F, having two des A B, BC of the one equal to two fides E D, F E of the other, each to each; and the angles A B C, FED contained under the equal fides are equal. [By 4. 1.] the base A c will be equal to the bafe D F, and the angle B A C equal to the angle FDE. But fince the angle ABE has been proved to be equal to the angle B E D, the outward angle BGC of the triangle ABG will be equal to the outward angle FH E of the triangle EH D. Therefore because the diagonal B'E cuts the diagonals A C, FD, fo that the alternate angles BG C, F HE are equal; [by 27. 1.] the diagonals A C, FD will be parallel. But they have been proved to be equal wherefore [by 33. 1.] the fides AF, CD joining them will be equal and parallel.

Secondly, I fay, the oppofite angles A, D, or F, C, are equal. For because ACDF has been proved to be a parallelogram, the oppofite angles FAC, FDC; AFD, ACD [by 34. 1.] will be equal. But fince the angles CAB, ACB of the triangle A B C are refpectively equal to the angles F DE, EFD of the triangle FE D, the correfpondent files and angles of these triangles being equal to one another; therefore will the angle S CAF, BA C be equal to the angles FDC, FDE; that is, the angle F A B equal to the angle CDE. In like manner the, angles AFD, EFD, as alfo the angles ACD, BCA will be equal; that is, the angle AFE equal to the angle B C D.

Thirdly, the diagonal B E divides the figure given into two equal parts.

For because [by 34. 1.] the diagonal BE of the parallelogram AB D E divides that parallelogram into two equal triangles A B E, E D B ; and because the triangles AFE, ECD having each of the fides of the one equal to each of the fides of the other, are alfo equal. Therefore the triangles AEB, AFE, as alfo EBD, BCD, will be equal; that is, the trapezium A B E F equal to the trapezium BCDE.

Therefore if there be any right-lined figure having fix fides, and two contiguous fides of it be alternately equal and parallel to the other contiguous and oppofite fides, each to each: I say, firft, the two remaining fides will be equal and parallel: Secondly, the oppofite angles (viz. the firft and fourth, fecond and fifth, third and fixth) will be equal to one another: Thirdly, any diagonal joining two of those F oppofite

oppofite angles, will divide the figure into two equal parts. Which was to be demonftrated.

SCHOLIU M.

Hence a hexagon, whofe oppofite fides are all alternately parallel, may be called a parallelogram, that is, a figure having parallel fides; as well as the quadrilateral figure mentioned by Euclid at prop. 33. lib. 1. and, indeed, fo may any other figure, having parallel fides, whatever be their number. Moreover, what Euclid has demonftrated at prop. 33, 34. lib. 1. of his four-fided parallelograms, holds univerfally true of any parallelogram of fix or more oppofite fides, when thefe fides are alternately equal aud parallel. Alfo, if the fides of any right-lined figure be alternately parallel, thofe that are oppofite will be equal; and all the diagonals bifect one another in one and the fame point. There are other uniform properties of thefe parallelograms, which I leave to others to discover and demonftrate.

PROP. VIII. PROB L. To make a triangle equal to a given right-lined

PQ

figure.

Let the given right-lined figure be ABCDEF: it is

B

required to make

a triangle equal to the given right-lin❜d figure

ABCDEF.

From one angle c, draw the diagonal c A to the angle A next but one to the angle c. Continue out the fide FA of the figure to G, being the firft oppofite fide to the angle c;. and from the angle B, lying

between

between the angles A and C, draw the right line B G [by 31. 1.] parallel to the diagonal CA, cutting the continuation of AF in the point G, and draw the right line CG, which let cut the fide A B of the figure in the point K. Again, draw a fecond diagonal CF of the figure; and continuing out the next fide F E of the figure to H, from the point G draw [by 31. 1.] the right line G H parallel to the second diagonal CF, meeting the continuation of the fide EF of the figure in the point H, and draw the right line C H from the angle c of the figure. Let this meet the right line GF in the point L. Again, draw a third diagonal CE of the given figure, and continuing out the fide FE of the figure to the point I, from the angle D draw the right line DI parallel to the diagonal c E, meeting the continuation of the fide F E of the figure in the point 1. And from the angle c draw the right line CI, meeting the fide DE of the figure in the point M: I fay, the triangle HCI will be equal to the given right-lined figure ABCDEF.

For becaufe [by conft.] the right line BG is parallel to the diagonal AC of the figure, the triangles CB G, GAB ftanding upon the fame bafe EG between these parallels, will [by 35. 1.] be equal to one another. And taking away from both the common triangle BG K, there will remain the triangle AKG equal to the triangle BKC. Therefore the right-lined figure AKC DEF, together with the triangle BK C, will be equal to the fame right-lined figure AKC DEF, together with the triangle AKG. Wherefore the right-lined figure GCDEF will be equal to the given right-lined figure A B C D E F.

Again, becaufe [by conft.] the right line G H is pa→ rallel to the diagonal CF, the triangles G H C, G F H between them standing upon the bafe G H, will be equal [by 35. 1.] and taking away the common triangle GLH, the triangles HLF, GLC will be equal; and fo the right-lined figure LFE DC, together with the triangle H L F, that is, the right-lined figure HE DC, will be equal to the fame right-lined figure LFE DC, together with the triangle GLC; that is, equal to the right-lined figure GCDEF, or to the given right-lined figure ABCDEF; because thefe two figures have already been proved to be equal to one another. After the fame manner we demonftrate, that the triangles DMC, EMI are equal to one another; and fo

F 2

the