ing angle ADC will alfo 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 also be equal to the fide AB, and the fide AD to the fide BC, and the two triangles will be equal in all refpects (Prop. 21.) Q. E. D.

PRO P. XXXI. THEOREM.

Parallelograms, and triangles, ftanding upon the fame base, 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 interfects 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 interfects 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

D 4

angle

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 ADF, BCE are mutually equiangular, and have two correfponding fides equal to each other, they will be equal in all refpects (Prop. 21.)

If, therefore, from the whole figure ABED, there be taken the triangle BCE, there will remain the parallelogram BD; 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 same thing, the remainders will be equal; confequently, the parallelogram AE is equal to the parallelogram BD..

Again, let ABC, ABF be two triangles, ftanding upon the fame base 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 AF (Prop. 28.)

Then, fince BD, AE, are two parallelograms, ftanding upon the fame bafe AB, and between the fame parallels AB, DE, they are equal to each other (Prop. 32.)

And, because the diagonals AC, BF bisect them (Prop. 31.), the triangle ABC will also be equal to the triangle ABF. Q. E. D.

PROP. XXXII. THEOREM.

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

Let the parallelogram AC and the triangle AEB ftand upon the same 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 AEB, because they stand upon the same base AB, and between the fame parallels AB, DE (Prop. 32.); whence the parallelogram AC is also double the triangle AEB. QE. 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.

PROP. XXXIII. PROBLEM.

To make a parallelogram that shall have its oppofite fides equal to two given right lines, and one of its angles equal to a given rectilineal angle.

Let AB and c be two given right lines, and D a given rectilineal angle; it is required to make a parallelogram that fhall have its oppofite fides equal to AB and c, and one of its angles equal to D.

At the point A, in the line AB, make the angle BAF equal to the angle D (Prop. 20.) and the fide AF equal to c (Prop. 3.)

Alfo, make FE parallel and equal to AB (Prop. 28 and 3.), and join BE; then will AE be the parallelogram required.

For, fince FE is parallel and equal to AB (by Const.), BE will be parallel and equal to AF (Prop. 30.); whence the figure AE is a parallelogram.

And, because AF is equal to c (by Conft.) BE will alfo be equal to c; and the angle BAF was made equal to the angle D.

The oppofite fides of the parallelogram AE are, therefore, equal to the two given lines AB and C; and one of its angles is equal to the given angle D, as was to be done.

PROP. XXXIV. THEOREM.

If two fides of a triangle be bifected, the right line joining the points of bifection, will be parallel to the bafe, and equal to one half

of it.

B

Let ABC be a triangle, whose fides CA, CB are bifected in the points D, E; then will the right line DE, joining thofe points, be parallel to AB, and equal to one half of it. For, in DE produced, take EF equal to ED (Prop. 3.), and join BF:

Then, fince EC is equal to EB (by Hyp.), ED to EF (by Conft.) and the angle DEC to the angle BEF (Prop. 15.), the fide BF will also be equal to the fide DC, or its equal DA, and the angle EFB to the angle EDC (Prop. 4.)

And, because the right line DF interfects the two right lines CD, FB, and makes the angle EDC equal to the alternate angle EFB, BF will be parallel to DC or DA (Prop. 24.)

The right lines BF, AD, therefore, being equal and parallel, the lines DF, AB, joining their extremes, will alfo be equal and parallel (Prop. 30.)

But DF is the double of DE (by Conft.); confequently AB is also the double of DE; that is DE is the half of AB.