For, fince the angles EGB, GHD are equal to each other (by Hyp.), and the angles AGH, EGB are alfo equal to each other (Prop. 15.), the angle AGĦ 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, thofe 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, alfo, equal to two right angles, (Prop. 13.) the angles agh, BGH will be equal to the angles BGH, GHD (Áx. 1.)

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

But these are alternate angles: therefore, in this cafe, 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.

Z

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

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, since 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 same manner, it may be shewn that it is not lefs; consequently they must be equal to each other. Q.E. D.

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

PRO P. 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.

H

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 Er intersects 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 oppofite 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, alfo, be equal to two right angles. D

Q.E.D.

COROLL.

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, nor less than it; confequently it must be greater, Q. E. D,

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

B

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. S.), and join BD.

Then, because CD is equal to CB (by conft.), the angle CRB 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 conf.); 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 fhall be equal to three given right lines, provided any two of them, taken together, be greater than the third.

B

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 whose 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, describe the circle KHD, cutting DG in K; and from the point F, with the distance FG, defcribe the circle GHL, cutting DG in L.

Then, becaufe 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