will be one right angle; that is, the angles F AC, AFB, BF C will be one right angle (for AFC is equal to AFB, BFC). In like manner it is proved, that the angles CFD, C D F are equal to one right angle. Therefore the angles FAC, AFB, BFC are equal to the angles CFD, CDF. And if the angle B F C be added in common, the angles F A C, A FB, 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 AFB, B F D be taken away, there will remain the angle FA C, and twice the angle B F C, equal to the angle CDF. Therefore the angle CDF exceeds the angle FA C by twice the angle BFC; and so 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 ADF of the triangle AF 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 DFC; or to the angles F AC, twice the angle B F D, and twice the angle DFC; or equal to the angle FA C, and twice the angle B FC. Therefore the angle ADF exceeds the angle FA C, by twice the angle B FC; and so the angle в г с 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 base upon it, and the right line bisecting that angle, will be one half the difference of the angles at the base. Which was to be demonftrated.

SCHOLIUM.

When the perpendicular F C falls without the triangle, it is evident, that the vertical angle AFD is the difference of the angles AFC, DFC at the perpendicular. But when the perpendicular F C falls within, the angle B F C itself. (which is one half the difference of the angles at the base) is also one half the difference of the angles CFD, AFD at the perpendicular. For because the angles AF B, BFD are [by Juppofition equal; if the angle BFC be added to both, the angle AFC will be equal to the angles BFD, BFC; that is, to the angle CDF, and twice BFC: wherefore the angle BFC is one half the difference of the angles AFC, CFD.

PROP.

PROP. VII. THEO R.

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; secondly, the opposite angles (viz. the first and fourth, second and fifth, third and fixth) will be equal to one another; thirdly, any diagonal joining two of those opposite angles, will divide the figure into two equal parts.

B

A

G

H

F

D

E

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

divide the figure into two equal parts.

For draw the diagonals AE, AC, BF, BD, CE, DF; and let the diagonal B E cut the diagonals AC, DF in the points G, H.

Then because the fides AB, ED are equal and parallel [by 33. 1.] the figure ABDE will be a parallelogram, having its fides AE, B D equal and parallel, and the angles ABE, BED equal. In like manner, because the fides в с, E F are equal, the figure FBCE will be a parallelogram, having the oppofite fides FB, ECc equal and parallel, and the angles FEB, CBE equal. Therefore the angles A B E, E B C will be equal to the angles BED, FEB; that is, the whole angle A B C equal to the whole angle FED. Where

fore fince there are two triangles A B C, DEF, having two des AB, BC of the one equal to two fides E D, FE of the other, each to each; and the angles ABC, FED contained under the equal fides are equal. [By 4. 1.] the base A c will be equal to the base DF, and the angle B A C equal to the angle FDE. But fince the angle A B E has been proved to be equal to the angle B ED, the outward angle BGC of the triangle ABG will be equal to the outward angle FH E of the triangle E HD. Therefore because the diagonal B'E cuts the diagonals A C, FD, so that the alternate angles BGC, FHE are equal; [by 27. 1.] the diagonals AC, FD will be parallel. But they have been proved to be equal: wherefore [by 33. 1.] the fides AF, C D 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 opposite angles FAC, FDC; AFD, ACD [by 34. 1.] will be equal. But fince the angles CAB, ACB of the triangle A B C are respectively equal to the angles FDE, EFD of the triangle FED, the correfpondent fides and angles of these triangles being equal to one another; therefore will the angle SCAF, BAC 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 AF E equal to the angle B CD.

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

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

Therefore if there be any right-lined figure having fix sides, and two contiguous fides of it be alternately equal and parallel to the other contiguous and opposite fides, each to each: I say, first, the two remaining fides will be equal and parallel: Secondly, 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 those oppofite

F

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

SCHOLIU M.

Hence a hexagon, whose opposite 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, so may any other figure, having parallel fides, whatever be their number. Moreover, what Euclid has demonstrated at prop. 33, 34. lib. 1. of his four-fided parallelograms, holds univerfally true of any parallelogram of fix or more opposite fides, when these Jides are alternately equal aud parallel. Also, if the fides of any right-lined figure be alternately parallel, those that are oppofite will be equal; and all the diagonals bifect one another in one and the same point. There are other uniform properties of these parallelograms, which I leave to others to discover and demonstrate.

PROP. VIII. PROB L.

To make a triangle equal to a given right-lined figure.

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

C

required to make a triangle equal to the given right-lin'd figure

ABCDEF.

D

M

B

K

E

From one angle c, draw the diagonal C A to

A

the angle A next but one to the

G

L

H

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 AB of the figure in the point k. Again, draw a second diagonal CF of the figure; and continuing out the next fide FE of the figure to H, from the point & draw [by 31. 1.] the right line G H parallel to the fecond diagonal CF, meeting the continuation of the fide EF of the figure in the point H, and draw the right line cu 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 side FE of the figure in the point 1. And from the angle c draw the right line c 1, meeting the fide DE of the figure in the point M: I say, the triangle HCI will be equal to the given right-lined figure ABCDEF.

For because [by conft.) the right line BG is parallel to the diagonal A C of the figure, the triangles CBG, GAB standing upon the same base B G between these parallels, will [by 35. 1.] be equal to one another. And taking away from both the common triangle BGK, there will remain the triangle AKG equal to the triangle в к.с. Therefore the right-lined figure AKCDEF, together with the triangle в кс, will be equal to the fame right-lined figure AKCDEF, together with the triangle AKG. Wherefore the right-lined figure GCDEF will be equal to the given right-lined figure A B CDEF.

Again, because [by conft.] the right line GH is parallel to the diagonal cr, the triangles GHC, GFH between them standing upon the base GH, will be equal [by 35. 1.] and taking away the common triangle GLH, the triangles HLF, GLC will be equal; and so the right-lined figure LFEDC, together with the triangle HLF, that is, the right-lined figure HEDC, will be equal to the fame right-lined figure LFEDC, together with the triangle GLC; that is, equal to the right-lined figure GCDEF, or to the given right-lined figure ABCDEF; EF; because these 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