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And, for the same 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 er, and the altitude DK to the altitude 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 since the diagonals de, 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 standing upon equal bases, and between the same parallels, are equal to each other.

PRO P. VI. THEOR E M.

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, since ac is a parallelogram, whose 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 dok will be equal to the triangle Dek, and the triangle KFB to the triangle KHB (1. 30.)

But, since 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.

PRO P. VII. THEOREM.

Parallelograms which are about the diagonal of a square are themselves squares.

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Let Bd be a square, and HE, FG parallelograms about its diagonal Ac; then will those parallelograms also be squares.

For since the side of the square AB is equal to the side 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 consequently the fide an is equal to the side AK (I. 6.)

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But the fide is equal to the side er, and the fide Hķ to the side AE (1. 30.); whence the figure he is equilateral.

It has also all its angles right angles :

For EAH is a right angle, being the angle of a square ; and HG, EF are each of them parallel to the fides of the fame square, whence the remaining angles will also be right angles (I. 25.)

The figure He, therefore, being equilateral, and hav. ing all its 'angles right angles, is a square : and the fame may be proved of the figure FG.

Q. E. D.

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The rectangles contained under a given line and the several parts of another line, any how divided, are, together, equal to the rectangle of the two whole lines,

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Let A and Bc be two right lines, one of which, BC, is divided into several parts in the points D, E; then will the rectangle of A and BC, be equal to the sum of the rectangles of A and BD, A and De, and A and ec.

For make BF perpendicular to BC (I. 11.) and equal to A (I. 3.), and draw FG parallel to Bc, and DH, ei

and cg each parallel to BF (I. 27.), producing them till they meet FG in the points H, I, G.

Then, since the rectangle Bh is contained by BD and BF (II. Def. 3.), it is also contained by BD and A, because BF is equal to A (by Conft.)

And, since the rectangle di is contained by de and DH, it is also contained by de and A, because ph is equal to BF (I. 30.), or A.

The rectangle eg, in like manner, is contained by EC and A; and the rectangle BG by BC and A.

But the whole rectangle Bc, is equal to the rectangles BH, di and Eg, taken together; whence the rectangle of A and Bc is also equal to the rectangles of A and BD, A and de and A and Ec, taken together. Q. E. D.

PRO P. IX. THEOREM.

If a right line be divided into any two parts, the rectangles of the whole line and each of the parts, are, together, equal to the square of the whole line.

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Let the right line AB be divided into any two parts in the point c; then will the rectangle of AB, AC, together with the rectangle of AB, BC, be equal to the square

of AB.

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For, upon AB describe the square AD (II. 1.), and through c draw of parallel to AE or BD (I. 27.)

Then, since the rectangle AF is contained by AE, AC, it is also contained by AB, AC, because AE is equal to AB (II. Def. 2.)

And, since the rectangle co is contained by BD, BC, it is also contained by AB, BC, because Bp is equal to AB,

But AD, or the square of AB, is equal to the rectangles AF, CD, taken together; whence the rectangle AB, AC, together with the rectangle AB, BC, is alfo equal to the square of AB.

Q. E, D,

PROP. X. THEOREM.

If a right line be divided into any two parts, the rectangle of the whole line and one of the parts, is equal to the rectangle of the two parts, together with the square of the aforesaid part.

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Let the right line AB be divided into any two parts in the point c; then will the rectangle of AB, BC be equal to the rectangle of AÇ, CB, together with the square of CB.

For upon ce describe the square CE (II. 1.), and through a draw AF parallel to CD (I. 27.), meeting ED, produced, in F,

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