And, because AB is equal to DC (by Hyp.), AC com. mon to each of the triangles ABC, ADC, and the angle DCA equal to the angle CAB, the fide AD will also be equal to the fide BC, and the angle DAC to the angle AÇв (Prop.4.)

Since, therefore, the right line AC interfects the two right lines AD, BC, and makes the alternate angles equal to each other, thofe lines will be parallel (Prop. 23.)

But the line AD has been proved to be equal to the line BC; confequently they are both equal and parallel. Q.E.D.

PROP. XXX. THEOREM.

The oppofite fides and angles of any paral lelogram are equal to each other, and the diagonal divides it into two equal parts,

Let ABCD be a parallelogram, whofe diagonal is AC ; then will its opposite fides and angles be equal to each other, and the diagonal AC will divide it into two equal parts.

For, fince the fide AD is parallel to the side BC (Def, 22.), and the right line AC interfects them, the angle DAC will be equal to the alternate angle ACB (Prop. 24.)

And, because the fide DC is parallel to the fide ab (Def. 22.), and AC intersects them, the angle DCA will be equal to the alternate angle CAB (Prop. 24.)

Since, therefore, the two angles DAC, DCA, are equal to the two angles ACB, CAB, each to each, the remaining

angle ADC will also be equal to the remaining angle ABC (Prop, 29. Cor) and the whole angle DAB to the whole angle DCB.

But, the triangles CDA, ABC, being mutually equiangular, and having AC common, the fide DC will alfo be equal to the fide AB, and the fide AD to the fide BC, and the two triangles will be equal in all respects (Prop 21.) Q.E.D.

PROP. XXXI. THEOREM,

Parallelograms, and triangles, ftanding upon the fame bafe, and between the fame parallels, are equal to each other.

Let AE, BD be two parallelograms ftanding upon the fame base AB, and between the fame parallels AB, DE; then will the parallelogram AE be equal to the parallelogram BD.

For, fince AD is parallel to BC (Def. 22.), and DE intersects them, the outward angle ECB will be equal to the inward oppofite angle FDA (Prop. 25.)

And, because AF-is parallel to BE (Def. 22.), and DE intersects them, the outward angle AFD will be equal to the inward oppofite angle BEC (Prop. 25.)

Since, therefore, the angle ECB is equal to the angle FDA, and the angle AFD to the angle BEC, the remaining

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 same 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.

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. 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 ECB has been fhewn to be equal to the angle GHD; therefore, the angles BGH, GUD, taken together, will, alfo, 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 lefs 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 GH interfects them, the angle AGH is equal to the alternate angle CHF (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 co, as was to be fhewn.