But fince a part of the circle CHL falls without the circle RHD, 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 fally without each other.

They muft, therefore, cut one another, in fome point H; and if the right lines EH, H 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, becaufe F is the centre of the circle GHL, FH is equal to FG (Def. 13.); but rG is equal to c (by Conft.); therefore FH is alfo equal to c.

And fince EE is, likewife, equal to B (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 rectilineal angle.

N

XX

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 shall be equal to BAC.

Take any point F in AB, and from the point A, at the distance AF, describe the circle FGS, cutting AC in Ġ; and join FG.

Make DK equal to AF, and KE equal to FG (Prop. 3.); and from the points D, K, at the diflances DK, KE, defcribe 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 alfo 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.

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 refpects.

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 also equal to the fide DF, the fide BC to the fide EF, and the two triangles will be equal in all respects.

For, if AC be not equal to DF, one of them muft 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.

The fide AC, therefore, cannot be greater than the fide DF; and, in the fame manner, it may be fhewn that it cannot be lefs; confequently it must be equal to it.

And, fince the two fides AC, AB, are equal to the two fides DF, DE, each to each, and the angle CAB is equal to the angle FDE, the fide BC will also be equal to the fide EF, and the two triangles will be equal in all refpects (Prop. 4.) Q, E.D.

PROP. XXII. THEOREM.

T..!

If a right line interfect two other right lines, and make the alternate angles equal to each other, those lines will be parallel.

Let the right line EF interfect the two right lines AB, CD, and make the alternate angles AEF, EFD equal to each other; then will AB be parallel to CD.

For, if they be not parallel, let them be produced, and they will meet each other, either on the fide AC, or on the fide BD (Def. 20.)

Suppofe them to meet in the point G, on the fide BD.. Then, fince FGE is a triangle, the outward angle AE is greater than the inward oppofite angle EFD (Prop. 16.) –

But

But the angles E3C, EBA are, together, equal to the angle ABC (Ax. 8.); confequently the angles ABC, ABD are, alfo, equal to two right angles. Q. E. D.

COROLL. All the angles which can be made, at any point B, on the fame side of the right line CD, are, together, equal to two right angles.

If a right line meet two other right lines, in the fame point, and make the angles on each fide of it together equal to two right angles, those lines will form one continued right line.

B

Let the right line AB meet the two right lines CB, BD, at the point в, and make the angles ABC, ABD together equal to two right angles, then will BD be in the fame right line with CB.

For, if it be not, let fome other line BE be in the fame right line with CB.

Then, because the right line AB falls upon the right Line CBE, the angles ABC, ABE, taken together, are equal to two right angles (Prop. 13.)

But the angles ABC, ABD are alfo equal to two right angles (by Hyp.); confequently the angles ABC, ABE are qual to the angles ABC, ABD.