For, fince the angles LGB, GHD are equal to each other (by Hyp.), and the angles AGH, EGB are alfo 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, fince 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, alfo, be parallel to CD (Prop. 22.) Q. E. D.

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

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

For if they be not equal, one of them must be greater than the other; let EFD be the greater; and make the angle EFB equal to AEF (Prop. 20.)

Then, fince AB, CD are parallel, the right line FB, which interfects CD, being produced, will meet AB in fome point в (Pof. 4.) ̈ ́

And, fince EFB is a triangle, the outward angle AEF will be greater than the inward oppofite angle EFB (Prop. 16.)

But the angles AEF, EFB are equal to each other (by Conft.) whence they are equal and unequal, at the fame time, which is abfurd.

The angle EFD, therefore, is not greater than the angle AEF; and, in the fame manner, it may be fhewn that it is not lefs; confequently they must be equal to each other. Q.E. D.

COROLL. Right lines which are perpendicular to one of two parallel right lines, are also perpendicular to the other.

PROP. XXV. THEOREM.

If a right line interfect two parallel right lines, the outward angle will be equal to the inward oppofite one, on the fame fide; and the two inward angles, on the fame fide, will be equal to two right angles.

お

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

For, fince the right line EF interfects the two parallel right lines AB, CD, the angle AGH will be equal to the alternate angle GHD (Prop. 24.)

But the angle AGH is equal to the opposite angle EGB (Prop. 15.); therefore the angle EGB will, alfo, be equal to the angle GHD.

Again, fince the right line BG falls upon the right line. EF, the angles EGB, BGH, taken together, are equal to two right angles (Prop. 13.)

But the angle EGB has been fhewn to be equal to the angle GHD; therefore, the angles BGH, GHD, taken together, will, also, be equal to two right angles. Q. E.D.

COROLL. If a right line intersect two other right lines, and make the two inward angles, on the fame fide, together less than two right angles, thofe lines, being produced, will meet each other.

PROP. XXVI. THEOREM.

Right lines which are parallel to the fame right line, are parallel to each other.

Let the right lines AB, CD be each of them parallel to EF, then will AB be parallel to CD.

For, draw any right line GK, cutting the lines. AB, EF, CD, in the points G, H and K.

Then, because AB is parallel to EF (by Hyp.), and GM interfects them, the angle AGH is equal to the alternate angle GHF (Prop. 24.)

And because CD is parallel to EF (by Hyp.), and нK intersects them, the outward angle GHF is equal to the inward angle HKD (Prop. 25.)

But the angle AGH has been fhewn to be equal to the angle GHF; therefore the angle AGK is alfo equal to the angle GKD.

And, fince the right line GK interfects the two right lines AB, CD, and makes the angle AGK equal to the alternate angle GKD, AB will be parallel to CD, as was to be fhewn.

PRO P. XXVII.

PROBLEM.

Through a given point, to draw a right line parallel to a given given right line.

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B

Let AB be the given right line, and c the given point ; it is required to draw a right line through the point c that shall be parallel to AB.

Take any point D in AB, and make DE equal to DC (Prop. 3.); and from the points C, E, with the distances. CD, ED, describe the arcs rs, nm.

Then, fince any two fides of the triangle ECD are, together, greater than the third fide (Prop. 18.), thofe arcs will interfect each other (Prop. 19.).

Let them interfect at F; and through the points F, C, draw the line GH, and it will be parallel to AB, as was required.

For, fince the fides CF, FE of the triangle EFC are each equal to the fide CD, or DE, of the triangle CDE, (by Conft.) and EC is common, the angle ECF will be equal to the angle CED (Prop. 7.)

But these are alternate angles; therefore GH is parallel to AB (Prop. 24.); and it is drawn through the point c. as was to be done.