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PRO P. XX. PROBLEM.

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

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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 that shall be equal to BAC.

Take any point F in AB, and from the point A, at the distance AF, describe the circle Fus, 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 distances DK, KE, deferibe the circles Klr and num, cutting each other in l.

Through the points d, I draw the right line on, and the angle EDN will be equal tó BAC, as was required.

For, join Kl: then since 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 bl (Ax. 1.); and Fg is equal to ke or KL (by Conft.)

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

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PRO P. XXI. THEOREM.

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

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Let the triangles ABC, der be' mutually equiangular, and have the side ab equal to the side de; then will the fide ac be also equal to the side DF, the side bc to the side er, and the two triangles will be equal in all respects.

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, since the two sides AB, AG, are equal to the two sides DE, DF, each to each, and the angle GAB is equal to the angle Fde (by Hyp.), the angle ABG will, also, be equal to the angle der (Prop. 4.)

But the angle DEF is equal to the angle ABC (by Hyp.); consequently the angle ABG will, also, be equal to the angle ABC, the less to the greates, which is absurd.

The side Ac, therefore, cannot be greater than the side DF; and, in the same manner, it may be fhewn that it cannot be less ; consequently it must be equal to it.

And, since 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 side bc will also be equal to the fide EF, and the two triangles will be equal in all respects (Prop. 4.) Q. E. D.

PRO P. XXII. THEOREM.

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

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

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

Suppose them to meet in the point G, on the side BD.

Then, fince Fre is a triangle, the outward angle aer is greater than the inward opposite angle EFD (Prop. 16.)

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But the angles, AEF, EFD, are equal to each other (bag Hyp:); whence they are equal and unequal at the same time, which is absurd.

The lines AB, CD, therefore, cannot meet on the side BD; and, in the same manner, it may be shewn that they cannot meet on the side AC; consequently they must be parallel to each other (Def. 20.) Q. E. D.

Coroll. Right lines which are perpendicular to the same right line are parallel to each other.

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PRO P. XXIII. THEOREM.

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If a right line intersect two other right lines, and make the outward angle equal to the inward opposite one, on the fame fide; or the two inward angles, on the same side, together equal to two right angles, those lines will be parallel.

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Let the right line er interfect the two right lines AB, CD, and make the outward angle EGB equal to the inward angle GHD; or the two inward angles BGH, GHD together equal to two right angles; then will AB be parallel to CD.

For, since the angles EGB, GHD are equal to each other (by Hyp.), and the angles AGH, EGB are also equal to each other (Prop. 15.), the angle AGH will be equal to the angle GHD (Ax. 1.)

But when a right line interfects two other right lines, and makes the alternate angles equal to each other, those lines will be parallel (Prop. 22.); therefore AB is parallel to CD.

Again, since the angles BGH, GHD are, together, equal to two right angles (by Hyp.), and AGH, BGH are, also, equal to two right angles (Prop. 13.), the angles AGH, BGH will be equal to the angles BGH, GHD (Ax. 1.)

And, if the common angle BGH be taken away, the remaining angle AGH will be equal to the remaining angle GHD (Ax. 3.)

But these are alternate angles; therefore, in this case, AB will, also, be parallel to CD (Prop. 22.) Q. E. D.

PRO P. XXIV. THEOREM.

If a right line intersect two parallel right lines, it will make the alternate angles equal to each other.

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Let the right line EF intersect the two parallel right lines AB, CD; then will the angle AEF be equal to the alternate angle EFD.

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