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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 base.

PROP. I. PROBLEM.

Upon a given right line to describe a square,

Let AB be the given right line; it is required to describe a square upon it.

Make AD, BC, each perpendicular and equal to AB (I. i and 3.), and join Dc; then will ac be the square required

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

And because AD, BC are equal and paraliel, AB, DC will, also, be equal and parallel (1. 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.) ; consequently the angles DCB, ADC are, also, right angles.

The figure AC, therefore, being both équilateral and rectangular, is a square ; and it is described upon the line AB, as was to be done.

PRO P. II. THEORE M.

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

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

For draw the diagonals AC; EG:

Then, since the two fides AB, Bc are equal to the two sides EF, FG, each to each (by Hyp.), and the angle B 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 rh; and in the same manner it may be proved when the figures are squares.

Q. E. D. PROP. III. THEOREM.

The sides and diagonals of equal squares are equal to each other.

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Let BD, FH, be two equal squares ; then will the fide AB be equal to the side er, and the diagonal Ac to the diagonal EG.

For if AB, EF be not equal, one of them must be greater than the other ; let AB be the greater, and make BL, BK each equal to er or FG (I. 3.); and join LK.

Then, because BL is equal to FE, BK to FG, and the angle LBK to the angle Erg, being each of them right angles, the triangle BLK will be equal to the triangle FeG (I. 4.)

But the triangle Feg is equal to the triangle BAC, being each of them the balves of the equal squares FH, BD (1. 30.); whence the triangle Blk is also equal to the triangle Bac, the less to the greater, which is absurd.

The side AB, therefore, is not greater than the side EF ; and in the same manner it may be proved that it cannot be less ; consequently they are equal to each other.

And because AB is equal to er, BC to FG, and the angle ABC to the angle erg (1. 8.), the fide ac will also be equal to the side EG (I. 4.)

Q. E. D.

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PRO P. IV. THEOREM.

The square of a greater line is greater than the square of a less; and the greater square has the greater side.

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Let the right line AB be greater than the right line efs then will BD, the square of AB, be greater than fh, the square of EF.

For since AB is greater than EF, and bc than FG (by Hyp.), takė BK, a part of Ba, equal to EF, and Bl, a part of BC, equal to FG (I. 3.); and join KL.

Then, because BK is equal to Fe, BL to FG, and the angle KBL to the angle EFG (I. 8.), the triangle BLK will be equal to the triangle Fge (I. 4.)

But the triangle BCA is greater than the triangle BLK, whence it is also greater than the triangle FGE.

And since the square bp is double the triangle BCA, and the square FH is double the triangle FGE (I. 30.), the square BD will also be greater than the square FH. Again, let the square bd be greater than the square

then will the fide All be greater than the side EF. For if all be not greater than EF, it muft be either equal to it, or less; but it cannot be equal to it, for then the

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square BD would be equal to the square FH (II. 2.), which it is not.

Neither can it be less, for then the square bd would be less than the square FH (II. 4.), which it is not ; consequently AB is greater than er, as was to be shewn.

PRO P. V. THEOREM.

Parallelograms and triangles, having equal bases and altitudes, are equal to each other.

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Let AC, EG be two parallelograms, having the base AB equal to the base EF, and the altitude DK to the altitude HL; then will the parallelogram ac be equal to the parallelogram EG.

For upon AB, EF, produced if necessary, let fall the perpendiculars CM, GN (1. 12.)

Then, fince MD, NH are rectangular parallelograms, the fide dc is equal to the side KM, and the side HG to the side LN (I. 30.)

But dc is also equal to AB, and HG to EF (I. 30.) ; therefore KM is equal to AB, and in to EF.

And, fince AB is equal to EF (by Hyp.), KM will be equal to LN; and consequently the rectangle md is equal to the rectangle NH (II. 2.)

But the rectangle MD is equal to the parallelogram Ac, because they stand upon the same base Dc, and between the same parallels Dc, AM. E 2

And,

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