If a parallelogram and a triangle stand upon the fame bafe, and between the fame parallels, the parallelogram will be double the triangle.

Let the parallelogram AC and the triangle AB stand upon the fame bafe AB, and between the fame parallels AB, DE; then will the parallelogram AC be double the triangle AEB.

For join the points B, D; then will the parallelogram AC be double the triangle ADB, because the diagonal DB divides it into two equal parts (Prop. 31.)

But the triangle ADB is equal to the triangle A EB, bẹcaufe they ftand upon the fame bafe AB, and between the fame parallels AB, DE (Prop. 32.); whence the parallelogram AC is alfo double the triangle AEB. Q.E. D.

COROLL. If the bafe of the parallelogram be half that of the triangle, or the bafe of the triangle be double that of the parallelogram, the two figures will be equal to each other.

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PROP.

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 AE will be greater than the inward opposite 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 shewn 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.

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 opposite angle GHD; and the two inward angles BGH, CHD will be equal to two right angles.

For, fince the right line Er 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 oppofite angle EGB (Prop. 15.); therefore the angle ECB 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 ECB has been fhewn to be equal to the angle GHD; therefore, the angles BGH, GUD, taken together, will, also, be equal to two right angles. Q. E.D.

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

PROP. XXVI. THEOREM.

Right lines which are parallel to the same 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 GĦ interfects them, the angle AGH is equal to the alternate angle CHF (Prop. 24.)

And because CD is parallel to EF (by Hyp.), and Hк 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 K 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. PROBLÉM,

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

I

Let AB be the given right line, and c the given point; it is required to draw à right line through the point c that fhall 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, defcribe 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.