angle CBE will be equal to the remaining angle DAF (Prop. 29. Cor. 1.)

But the fide AD is alfo equal to the fide BC (Prop. 31.) ' confequently, fince the triangles ADE, BCE are mutually equiangular, and have two correfponding fides equal to each other, they will be equal in all respects (Prop. 21.)

If, therefore, from the whole figure ABED, there be taken the triangle BCE, there will remain the parallelogram BP; and if, from the fame figure, there be taken the triangle ADF, there will remain the parallelogram AE.

But if equal things be taken from the fame thing, the remainders will be equal; confequently, the parallelogram AE is equal to the parallelogram BD,

Again, let ABC, ABF be two triangles, standing upon the fame bafe AB, and between the fame parallels, AB, CF; then will the triangle ABC be equal to the triangle

ABF.

For produce CF, both ways, to D and E, and draw AD parallel to BC, and BE to Ar (Prop. 28.)

Then, fince BD, AE, are two parallelograms, ftanding upon the fame base AB, and between the fame parallels AB, DE, they are equal to each other (Prop.

32.)

And, because the diagonals AC, BF bifect them (Prop. 51.), the triangle ABC will alfo be equal to the triangle ABF. Q.E.D.

PROP. XXXII.

THEOREM.

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

Let the parallelogram AC and the triangle AFB ftand upon the fame base 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 base of the parallelogram be half that of the triangle, or the base of the triangle be double that of the parallelogram, the two figures will be equal to each other.

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 (Pos. 4.)

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

But the angles. AEF, EFB are equal to each other (by Conft.) whence they are equal and unequal, at the same 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 alfo 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.

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 interfects the two parallel right lines AB, CD, the angle AGH will be equal to the alternate angle GHD (Prop. £4.)

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, also, be equal to two right angles.

D

Q. E. D.

COROLL.

COROLL. If a right line interfect two other right lines, and make the two inward angles, on the fame fide, together less than two right angles, those 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 GH intersects them, the angle AGH is equal to the alternate angle GHF (Prop. 24.)

And because CD is parallel to EF (by Hyp.), and HK interfects 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.