PROP. XXXII. THEOREM.

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

E

Let the parallelogram AC and the triangle ALB 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, bę cause they ftand upon the fame base 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 base of the triangle be double that of the parallelogram, the two figures will be equal to each other.

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,

F

Let AB and C be two given right lines, and D a given rectilineal angle; it is required to make a parallelogram that shall 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 base, and equal to one half of it.

Let ABC be a triangle, whose fides CA, CB are bifected in the points D, E; then will the right line DE, joining those 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 pF 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 alfo the double of DE; that is DE is the half of AB.

Q.E.D.

PROP.

PROP. XXXV,

PROBLEM,

To divide a given finite right line into any propofed number of equal parts,

Let AB be the given right line; it is required to divide it into a certain propofed number of equal parts.

From the point A, draw any right line AC, in which take the equal parts AD, DE, EC, at pleasure, (Prop. 3.) to, the number proposed,

Join BC; and parallel thereto draw the right lines EF, DG, (Prop. 28.) cutting AB, in F and G; then will ab be divided into the fame number of equal parts with AC, as was required.

For take EH, CK, each equal to DG (Prop. 3.), and join D, H and E, K.

Then, fince DG is parallel to EF (by Conft.), and ae interfects them, the outward angle ADG will be equal to the inward oppofite angle DEH (Prop. 25.)

And, because the fides AD, DG of the triangle AGD, are equal to the fides DE, EH of the triangle DHE (by Conft.), and the angle ADG is equal to the angle DEH, the base AG will also be equal to the bafe DH, and the angle DAG to the angle EDH (Prop. 4.)

But, fince the right line AE interfects the two right lines, AG, DI, and makes the outward angle EDH equal

to the inward oppofite angle DAG, DH will be parallel to AG or GF (Prop. 23.)

And, in the fame manner it may be shown, that EK is equal to AG, and parallel to AG or FB.

A

The figures GH, FK, therefore, being parallelograms, the fide DH will be equal to the fide GF, and the fide EK to the fide FB (Prop. 31.)

But DH, EK have been each proved to be equal to AG; confequently GF, FB are, also, each equal to AG; whence the line AB is divided into the fame number of equal parts with AC, as was to be done.