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1. The exterior angle BGC is equal to the interior and opposite

angle BDA (I. 29),

H

F

E

but

2. The angle BDA is equal to the angle DBA (I. 5), because BA is equal to AD, being sides of a square ; wherefore

3. The angle BGC is equal to the angle DBA or GBC; and therefore

4. The side BC is equal to the side CG (I. 6), but BC is equal also to GK, and CG to BK (I. 34); wherefore

5. The figure CGKB is equilateral. It is likewise rectangular ; for, since CG is parallel to BK, and BC meets them, therefore

6. The angles KBC, BCG are equal to two right angles (I. 29), but the angle KBC is a right angle (Def. 30, constr.), wherefore

7. BCG is a right angle ; and therefore also the angles CGK, GKB, opposite to these, are right angles (I. 34), wherefore

8. The figure CGKB is rectangular. But it is also equilateral, as was demonstrated, wherefore

9. CGKB is a square, and it is upon the side CB. For the same reason,

10. HF is a square, and it is upon the side HG, which is equal to AC (I. 34). Therefore

11. The figures HF, CK are the squares on AC, CB. And because the complement AG is equal to the complement GE (I. 43), and that

12. AG is the rectangle contained by AC, CB;

for GC is equal to CB; therefore also

13. GE is equal to the rectangle AC, CB; wherefore

14. AG, GE are equal to twice the rectangle AC, CB; and HF, CK are the squares on AC, CB; wherefore

15. The four figures HF, CK, AG, GE, are equal to the squares

on AC, CB, and twice the rectangle AC, CB; but HF, CK, AG, GE make up the whole figure ADEB, which is the square on AB; therefore

16. The square on AB is equal to the squares on AC, CB, and

twice the rectangle AC, CB.

Wherefore, if a straight line be divided, &c. Q.E.D. Cor. From the demonstration it is manifest that the parallelograms about the diameter of a square are likewise squares.

PROPOSITION 5.-Theorem.

If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D.

Then the rectangle AD, DB, together with the square on CD, shall be equal to the square on CB.

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Construction. Upon CB describe the square CEFB (I. 46) : join BE; through D draw DHG parallel to CE or BF (I. 31), meeting BE in H, and EF in G; and through H draw KLM parallel to CB or EF, meeting CE in L, and BF in M; also through A draw AK parallel to CL or BM, meeting MLI in K.

Demonstration. Then because the complement CH is equal to the complement HF (I. 43), to each of these equals add DM ; therefore

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1. The whole CM is equal to the whole DF; but because the line AC is equal to CB, therefore

2. AL is equal to CM (I. 36); therefore also

3. AL is equal to DF; to each of these equals add CH; and therefore

4. The whole AH is equal to DF and CH; but AH is the rectangle contained by AD, DB; for DH is equal to DB, and DF together with CH is the gnomon CMG; therefore

5. The gnomon CMG is equal to the rectangle AD, DB: to each of these equals add LG, which is equal to the square on CD (II. 4. Cor.), therefore

6. The gnomon CMG, together with LG, is equal to the rect

angle AD, DB, together with the square on CD; but the gnomon CMG and LG make up the whole figure CEFB, which is the square on CB; therefore

7. The rectangle AD, DB, together with the square on CD, is equal to the square on CB.

Wherefore, if a straight line, &c. Q.E.D. Cor. From this proposition it is manifest that the difference of the squares on two unequal lines AC, CD, is equal to the rectangle contained by their sum (AD) and their difference (DB).

PROPOSITION 6.-Theorem.

If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D.

Then the rectangle AD, DB, together with the square on CB, shall be equal to the square on CD.

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Construction. Upon CD describe the square CEFD (I. 46), and join DE; through B draw BHG parallel to CE or DF (I. 31), meeting DE in H, and EF in G; through H draw KLM parallel to AD or EF, meeting DF in M, and CE in L; and through A draw AK parallel to CL or DM, meeting MLK in K.

Demonstration. Then because the line AC is equal to CB, therefore

1. The rectangle AL is equal to the rectangle ti (I. 36), but CH is equal to HF (I. 43), therefore

2. AL is equal to HF; to each of these equals add CM; therefore

3. The whole AM is equal to the gnomon CMG; but AM is the rectangle contained by AD, DB, for DM is equal to DB (II. 4. Cor.), therefore

4. The gnomon CMG is equal to the rectangle AD, DB; to each of these equals add LG, which is equal to the square on CB, therefore

5. The rectangle AD, DB, together with the square on CB, is

equal to the gnomon CMG, and the figure LG; but the gnomon CMG and LG make up the whole figure CEFD, which is the square on CD, therefore

6. The rectangle AD, DB, together with the square on CB, is equal to the square on CD.

Wherefore, if a straight line, &c. Q.E.D.

PROPOSITION 7.-Theorem,

If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part.

Let the straight line AB be divided into any two parts in the point C.

Then the squares on AB, BC shall be equal to twice the rectangle AB, BC, together with the square on AC.

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Construction. Upon AB describe the square ADEB (I. 46), and join BD; through draw CF parallel to AD or BE (I. 31), meeting BD in G, and DE in F; through G draw HGK parallel to AB or DE, meeting AD in H, and BE in K.

Demonstration. Then because AG is equal to GE (I. 43), add to each of them CK; therefore

1. The whole AK is equal to the whole CE, and therefore

2. AK, CE are double of AK; but AK, CE are the gnomon AKF and the square CK; therefore

3. The gnomon AKF and the square CK are double of AK; but

4. Twice the rectangle AB, BC, is double of AK; for BK is equal to BC (II. 4. Cor.), therefore

5. The gnomon AKF and the square CK are equal to twice

the rectangle AB, BC: to each of these equals add HF, which is equal to the square on AC; therefore

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