PRO P. 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.

A

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 DF intersects 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.

J

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, DH, 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.

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

:1

BOOK

BOOK II

DEFINITIONS

1. A rectangle is a parallelogram whose angles are all right angles.

2. A fquare is a rectangle, whose fides are all equal to each other.

3. Every rectangle is said to be contained by any two of the right lines which contain one of the right angles.

4. If two right lines be drawn through any point in the diagonal of a parallelogram, parallel to its oppofite Gides, the figures which are intersected by the diagonal are called parallelograms about the diagonal.

5. And the other two parallelograms, which are not interfected by the diagonal, are called complements to the parallelograms which are about the diagonal.

田

6. In every parallelogram, either of the two parallelograms about the diagonal, together with the two complements, is called a gnomon.

7. The altitude of any figure is a perpendicular drawn from the vertical angle to the base.

UPON a given right line to describe a square.

Let AB be the given right line; it is required to defcribe a fquare upon it.

Make AD, BC, each perpendicular and equal to AB (I. 11 and 3.), and join DC; then will AC be the fquare required.

For, fince the angles DAB, ABC are right angles (by Conft.), AD will be parallel to BC (I. 22 Cor.)

And because AD, BC are equal and parallel, AB, DC

will, alfo, be equal and parallel (I. 30.)

But AD, BC are each equal to AB (by Conft.); whence AD, AB, BC and CD are all equal to each other.

The