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For, if AB be not greater than AC, it must be either equal or lefs.

But it cannot be equal, for then the angle ACB would be equal to the angle ABC (Prop. 5.), which it is not. Neither can it be lefs, for then the angle ACB would be lefs than the angle ABC (Prop. 17.), which it is not.

The fide AB, therefore, is neither equal to AC, noṛ lefs than it; consequently it must be greater. Q. E. D.

PROP. XVIII. THEOREM.

Any two fides of a triangle, taken together, are greater than the third fide.

A

Let ABC be a triangle; then will any two fides of it, taken together, be greater than the third fide.

For, in AC produced, take CD equal to CB (Prop. 3.), and join BD.

Then, because CD is equal to CB (by conft.), the angle CDB will be equal to the angle CBD (Prop. 5.)

But the angle ABD is greater than the angle CBD, confequently it must also be greater than the angle ADB.

And, fince the greater fide of every triangle, is oppofite to the greater angle (Prop. 17.), the fide AD is greater than the fide AB.

But AD is equal to AC and CB taken together (by conft.); therefore AC, CB are alfo greater than AB.

And, in the fame manner, it may be fhewn, that any other two fides, taken together, are greater than the third fide. Q. E. D.

PROP. XIX. PROBLEM.

To defcribe a triangle, whofe fides shall be equal to three given right lines, provided any two of them, taken together, be greater than the third.

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Let A, B, c be the three given right lines, any two of which, taken together, are greater than the third; it is required to make a triangle whofe fides fhall be equal to A, B, C respectively.

Draw any right line DG; on which take DE equal to A, EF equal to B, and FG equal to C (Prop. 3.)

From the point E, with the distance ED, defcribe the circle KHD, cutting DG in K; and from the point F, with the diftance FG, defcribe the circle GHL, cutting DG in L.

Then, because EG is greater than ED (by Hyp.), or its equal EK, the point G, which is in the circumference of the circle GHL, will fall without the circle KHD.

And, because FD is greater than FG (by Hyp.), or its equal FL, the point D, which is in the circumference of the circle KHD, will fall without the circle GHL.

But

But fince a part of the circle GHL falls without the circle KHD, and a part of the circle KHD falls without the circle GHL, neither of the circles can be included within the other.

Again, because DE, FG, or their equals EK, FL are, together, greater than EF (by Hyp.), the two circles can neither touch nor fall wholly without each other.

They muft, therefore, cut one another, in fome point H; and if the right lines EH, FH be drawn, EHF will be the triangle required.

For, fince E is the centre of the circle KHD, EH is equal to ED (Def. 13.); but ED is equal to a (by Conft.); therefore EH is alfo equal to A.

And, because F is the centre of the circle GHL, FH is equal to FG (Def. 13.); but FG is equal to c (by Conft.); therefore FH is alfo equal to c.

And fince EF is, likewife, equal to в (by Conft.), the three fides of the triangle EHF are refpectively equal to the three given lines A, B, C, which was to be fhewn.

PROP. XX. PROBLEM.

At a given point, in a given right line, to make a rectilineal angle equal to a given rectilincal angle.

A

m

N

Let DE be the given right line, D the given point, and BAC the given rectilineal angle; it is required to make an angle at the point D that fhall be equal to BAC.

Take any point F in AB, and from the point A, at the diftance AF, defcribe the circle FGS, cutting AC in G; and join FG.

Make DK equal to AF, and KE equal to FG (Prop. 3); and from the points D, K, at the diftances DK, KE, describe the circles KLr and nLm, cutting each other in L.

Through the points D, L draw the right line DN, and the angle EDN will be equal to BAC, as was required.

For, join KL then fince AG is equal to AF (Def. 13.), and AF is equal to DK (by Conft.), AG will also be equal to DK (Ax. 1.)

But DK is equal to DL (Def. 13.); confequently AG is also equal to DL (Ax. 1.); and FG is equal to KE or KL (by Conft.)

The three fides of the triangle DKL are, therefore, equal to the three fides of the triangle AFG, each to each;

whence

whence the angle KDL is equal to the angle FAG, or BAC (Prop. 7.); and it is made at the point D, as was to be done.

PROP. XXI. THEOREM.

If two triangles be mutually equiangular, and have two correfponding fides equal to each other, the other correfponding fides will also be equal, and the two triangles will be equal in all respects.

G

A A

B

Let the triangles ABC, DEF be' mutually equiangular, and have the fide AB equal to the fide DE; then will the fide AC be alfo equal to the fide DF, the fide BC to the fide EF, and the two triangles will be equal in all refpects.

For, if AC be not equal to DF, one of them must be greater than the other; let AC be the greater, and make AG equal to DF (Prop. 3.); and join BG.

Then, fince the two fides AB, AG, are equal to the two fides DE, DF, each to each, and the angle GAB is equal to the angle FDE (by Hyp.), the angle ABG will, alfo, be equal to the angle DEF (Prop. 4.)

But the angle DEF is equal to the angle ABC (by Hyp.); confequently the angle ABG will, alfo, be equal to the angle ABC, the lefs to the greater, which is abfurd.

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