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the triangle EM I, together with the right-lined figure HEM C, will be equal to the triangle C M D together with that figure; that is, the triangle HCI will be equal to the right-lined figure HEDC. But the right-lined figure HEDC has been proved to be equal to the right-lined figure GFEDC; and this laft right-lined figure, equal to the given right-lined figure ABCDEF. Therefore the triangle H CI will be equal to the given right-lined figure A B C D E F. Therefore, &c. Which was to be demonftrated.

SCHOLIU M.

This is one of the most elegant and useful problems in rightlined Geometry. Some land meafurers, who are used often to folve this problem upon paper, do it fo easily and expeditiously, by means of a parallel ruler and compaffes, as toob to be triangle required without actually drawing any diagonals at all, or any parallels to them.

PROP. IX. PROB L.

To make a parallelogram equal to a given triangle, whofe four fides fhall be equal to the three fides of the given triangle.

Let the given triangle be A B C: It is required to make a parallelogram equal to the given triangle A B C, having its four fides equal to the three fides of the given triangle.

Draw [by 31. 1.] the right line A M from the given point A parallel to the bafe B C of the triangle. And becaufe, if neither of the angles B, C be a right angle [by 19. 1.] each of the fides AB, AC is greater than a perpendicular drawn from the angles A,

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B, C, or the point D, to the oppofite parallel: But if any of the angles be a right angle, that is, either of the fides be perpendicular to the faid parallels, both the

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fides A B, A C of the

K H

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greater than twice the

frid perpendicular; and so half the fum of the fides will be

greater

greater than that perpendicular; that is, if the righ line GH be taken equal to the fide A в, and нi equ I to the fide A C, fo that G I he the fum of the fides A 3, A C, nd GI be divided at K into two cq a! rats; this half K will be greater than the perp ndicular draw. fom: D to AM: therefore bifect [by 9. 1.1 th bfe BC of the given triangle in the point D; and about the centre D, with a diftarce equal to G K or K 1, defcribe a circle cutting the parallight line AM in fom point, as L. Upon Aм take LM equal to B D, and join he points ", M. Then [by 33.1.] the figure BMLD is a parallelogram: I fav, the parallelogram B M L D is equal to the given triangle ABC, and the fum of its fides BM, ML LD, BD, will be equal to the fum of the fides A B, A C, BC of the given triangle.

For becaufe [by conft.] the right light line A M is parallel to the bafe B c of the triangle, and B M L D is a parallelogram upon B D, half the bafe of the triangle, having its oppofite fide M L in AM, the parallelogram BMLD will be [by 41. 1.] made equal to the given triangle ABC. Again, fince [by conft.] DL, BM are cach equal to G K; that is, to one half the fum of the fides A B, AC; and accordingly, the fum of DL, BM is equal to the fum of the fides AB, A c of the triangle; and the right lines B D, ML are equal to the bafe Bc of the triangle. Ther.fore the four fides DL, B M, B D, ML of the parallelogram BMLD will be equal to the three fides A B, AC, BC of the given triangle: but the parallelogram B M L D is allo equal to the given triangle ABC.

Therefore, &c. Which was to be done.

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In every triangle any parallelograms defcribed upon two fides are equal to the parallelogram defcribed upon the remaining fide, and whofe other fide is equal and parallel to the right line drawn from the angle contained under the fift-mentioned fides, to that point wherein the fides of the parallelograms oppofite to the fides of the triangle, when produced towards the angle, meet.

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Let there be any triangle ABC, and let any parallelograms A B D E, ACFG be defcribed upon the two fides A B, AC of it; whofe two fides D E, F G, oppofite to the aflumed fides AB, AC of the triangle produced towards the angle A, contained under the faid fides A B, A C, meet in the point H. Draw the right line A H: I fay, the parallelograms AD, AF are equal to the parallelogram defcribed upon the fide BC, and its other fide is equal and parallel to the right line A H. For continue out the right line HA to cut the fide Be of the triangle in the point L; and from the points B, C draw [by 31. 1.] the right lines B 1, CK parallel to the right line AH, and join the right line I K.

I

B

H

E

G

A

L

Then because [by fuppofition] K BIHA, CKHA are parallelograms; [by 34. I.] BI, CK will each of them be equal to A H; and fo BI will be equal to CK; and because [by 30. 1.] they are alfo parallel to one another, fince each of them is parallel to the right line A H; the right lines B C, IK will alfo be [by 33. 1.] parallel and equal. Therefore BCKI is a parallelogram upon в C, having its other fide BI or CK equal and parallel to the right line HA; which we are to prove to be equal to the parallelograms A D, A F. Therefore becaufe the parallelograms A D, ABIH are [by 35. 1.] equal; fince they ftand upon the fame bafe A B, and are between the fame parallels A B, HD; and [by 35. 1.] the parallelogram A B I H is equal to the parallelogram IL, because they have both the fame base BI, and stand between the fame parallels B I, LH; the parallelogram A D will be equal to the parallelogram IL. After the fame manner we demonftrate, that the parallelogram A F is equal to the parallelogram K L : therefore the parallelograms A D, A F, will be equal to the parallelogram B K.

Therefore in every triangle, any parallelograms described upon two fides are equal to the parallelogram defcribed upon the remaining fide, and whofe other fide is equal and parallel to the right line drawn from the angle contained

under

under the first-mentioned fides, to that point wherein the fides of the parallelograms oppofite to the fides of the triangle, when produced towards that angle, meet. Which was to be demonstrated.

SCHOLIU M.

This theorem is to be found in Pappus's mathematical collections. It is much more general than the 47th propofition of the first book of Euclid, which is only that particular cafe of this, where the parallelogram defcribed upon the two affumed fides of the triangle are each of them Squares, and the angle of the triangle contained under these two fides, is a right angle.

PROP. XI. PROBL.

To divide a given triangle into two equal parts, by a right line drawn from any given point in one of its fides.

Let there be a given triangle ABC, and D any given point in one of its fides: It is required to draw a right line from the given point D that shall divide the triangle ABC into two equal parts.

Bifect [by 10. 1.] the fide A C of the triangle in the point E, and join the right line EB; then if the point E falls in the point D, the thing is done; for [by 38. 1.) the triangles A BE, B C E are equal; and each of them will be one half of the given triangle ABC. Alfo if the given point be in either of the angles of the triangle, the thing is done by drawing a right line from that angle to the middle point of the oppofite fide.

In all other cafes, draw the right line EF [by 31. 1.] parallel to the right line D B, cutting the fide A B of the triangle in the point F, and join the right line D F.

Then becaufe [by A

conftr.] EF is parallel

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to the right line BD, the triangles F E D, F BE standing

F 4

upon

upon the fame base F E, between thofe parallels, will [by 35. 1.] be equal to one another; and adding to both the common triangle AFE, the triangles AFE, FB E will be equal to the triangles A FE, BFD; that is, the triangles ABE, AFD will be equal to one another. But it has been proved, that the triangle ABE is one half the given triangle ABC therefore alfo the triangle AFD will be one half of the given triangle ABC; and fo the remaining trapezium F BCD will be one half of the given triangle ABC. Wherefore the right line D F, drawn from the given point D in the fide of the given triangle ABC, will divide it into two equal parts. Which was to be done.

PROP. XII. PROB L.

To divide a given trapezium into two equal parts by a right line drawn from a given point in the middle of one of its fides,

B H

Let ABCD be a given trapezium, and E a given point in the middle of one of its fides: It is required to draw a right line from the given point E, that will divide the given trapezium into two equal parts. For [by 31. 1.] draw the right line F C from the angle c of the figure parallel to the fide A D, Cutting the fide AB in the C point F; bifect [by 10. 1.] the right line FC in the point G. And having drawn a right line from the given point E to the angle B of the figure, draw a right D line GH from G parallel to the right line E B, meeting the fide B C of the figure in the point H; and join the right line EH; and the business is done.

F

G

A

E

For join the right lines EF, EC, EG, BG.

Then because A E is equal to E D, and FC is parallel to AD [by conftr.] the triangles A FE, ECD will be equal to one another [by 37. 1.]. And because F G, GC are equal [by conftr.] the triangles F EG, GEC will [by 37. 1.] be equal to one another; as alfo the triangles FB G, GB.C. Therefore the triangles A FE, FEG will be equal

to

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