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PRO P. XXXV. PROBLEM.

To divide a given finite right line into any proposed number of equal parts.

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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. 1.) to the number propofed.

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 to 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 base DH, and the angle DAG to the angle EDH (Prop. 4.)

But, fince the right line AE interfects the two right lines DG, EF, 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 also 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.)

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.

BOOK

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DEFINITIONS.

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

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

3. Every rectangle is faid 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 fides, the figures which are interfected 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.

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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 bafe.

PROP. I. PROBLEM.

Upon a given right line to defcribe 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 paraliel, 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.

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The figure AC, therefore, is an equilateral parallelogram; and it has, likewise, all its angles right angles.

For the angle DAB is equal to the angle DCB, and the angle ABC to the angle ADC (I. 30.)

But the angles DAB, APC are right angles (by Conft.) ; confequently the angles DCB, ADC are, alfo, right angles. The figure AC, therefore, being both equilateral and rectangular, is a square; and it is defcribed upon the line AB, as was to be done.

PROP. II. THEOREM.

Rectangles and Squares contained under equal lines are equal to each other.

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Let BD, FH be two rectangles, having the fides AB, BC equal to the fides EF, FG, each to each; then will the rectangle BD be equal to the rectangle FH.

For draw the diagonals AC, EG:

Then, fince the two fides AB, BC are equal to the two fides EF, FG, each to each (by Hyp.), and the angle в is equal to the angle F (I. 8.), the triangle ABC will be equal to the triangle EFG (I. 4.)

But the diagonal of every parallelogram divides it into two equal parts (I. 30.); whence the halves being equal, the wholes will also be equal.

The rectangle BD is, therefore, equal to the rectangle FH; and in the fame manner it may be proved when the figures are fquares.

Q. E. D.

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