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PRO P. III. THEOREM.

The fides and diagonals of equal fquares are equal to each other.

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Let BD, FH, be two equal fquares; then will the fide AB be equal to the fide EF, 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 EF or FG (I. 3.); and join LK.

Then, because BL is equal to FE, BK to FG, and the angle LBK to the angle EFG, 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 halves of the equal fquares FH, BD (I. 30.); whence the triangle BLK is alfo equal to the triangle BAC, the less to the greater, which is abfurd.

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

And because AB is equal to EF, BC to FG, and the angle ABC to the angle EFG (I. 8.), the fide AC will also be equal to the fide EG (I. 4.) Q. E. D.

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PROP. IV. THEOREM.

The fquare of a greater line is greater than the fquare of a lefs; and the greater square has the greater fide.

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Let the right line AB be greater than the right line EF; then will BD, the fquare of AB, be greater than FH, the fquare of EF.

For fince AB is greater than EF, and BC than FG (by Hyp.), take вK, 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 alfo greater than the triangle FGE.

And fince the fquare BD is double the triangle BCA, and the fquare FH is double the triangle FGE (I. 30.), the fquare BD will also be greater than the square FH.

Again, let the fquare BD be greater than the square FH; then will the fide AB be greater than the fide EF. For if AB be not greater than EF, it must be either equal to it, or lefs; but it cannot be equal to it, for then the

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

Neither can it be less, for then the fquare BD would be less than the fquare FH (II. 4.), which it is not; confequently AB is greater than EF, as was to be fhewn.

PROP. V. THEOREM.

Parallelograms and triangles, having equal bafes 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 then will the parallelogram AC be equal to the parallelogram EG.

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

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

But DC is alfo equal to AB, and HG to EF (I. 30.); · therefore KM is equal to AB, and LN to EF.

And, fince AB is equal to EF (by Hyp.), Kм 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 ftand upon the fame base DC, and between the fame parallels DC, AM.

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And, for the fame reason, the rectangle NH is equal to the parallelogram EG; whence the parallelogram Ac is equal to the parallelogram EG.

Again, let ABD, EFH be two triangles, having the base AB equal to the base EF, and the altitude DK to the altitude HL; then will the triangle ABD be equal to the triangle EFH.

For, if the parallelograms AC, EG be compleated, they will be equal to each other, by the former part of the propofition.

And fince the diagonals DB, HF divide them into two equal parts (I. 30.), the triangle ABD will also be equal to the triangle EFH.

Q. E. D. COROLL. Parallelograms and Triangles ftanding upon equal bases, and between the fame parallels, are equal to each other.

PROP. VI. THEOREM.

The complements of the parallelograms which are about the diagonal of any parallelogram are equal to each other.

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Let AC be a parallelogram, and AK, KC, complements about the diagonal BD; then will the complement AK be equal to the complement KC.

For, fince AC is a parallelogram, whofe diagonal is BD, the triangle DAB will be equal to the triangle BCD (I. 30.)

And, because EG, HF are also parallelograms, whose diagonals are DK, KB, the triangle DGK will be equal to the triangle DEK, and the triangle KFB to the triangle KHB (I. 30.)

But, fince the triangles DGK, KFB are, together, equal to the triangles DEK, KHB, and the whole triangle DAB to the whole triangle DCB, the remaining part AK will be equal to the remaining part KC. Q. E. D.

PROP. VII. THEOREM.

Parallelograms which are about the diagonal of a fquare are themselves fquares.

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Let BD be a square, and HE, FG parallelograms about its diagonal AC; then will those parallelograms also be fquares.

For fince the fide of the fquare AB is equal to the fide BC, the angle CAB will be equal to the angle ACB (I. 5.) And because the right line GH is parallel to the right line CB, the angle AKH will also be equal to the angle ACB (I. 25.)

The angles CAB, AKH are, therefore, equal to each other; and confequently the fide AH is equal to the fide HK (I. 6.)

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