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to the squares on AC and CB, together with twice the rectangle AC. CB.

21. But HF, CK, AG, GE, make up the whole figure ADEB, which is the square on AB.

22. Therefore the square on AB is equal to the squares on AC and CB, and twice the rectangle AC.CB.

Conclusion.—Therefore, if a straight line, &c. Q. E. D.

Corollary.-It follows from this demonstration, that the parallelograms about the diagonal of a square are likewise squares.

PROPOSITION 5.-THEOREM.

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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.

(References-Prop. I. 31, 34, 36, 43, 46; II. 4, cor.) Hypothesis.—Let the straight line AB be bisected in C, and divided unequally in D.

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

Construction.—1. Upon CB describe the square CEFB (I. 46), and join BE.

2. Through D draw DHG parallel to CE or BF. (1. 31.)

3. Through H draw KLM parallel to CB or EF.

4. And through A draw AK parallel to CL or BM.

Demonstration.-1. Then the complement CH is equal to the complement HF. (I. 43.)

2. To each of these add DM; therefore the whole CM is equal to the whole DF. (ax. 2.)

3. But CM is equal to AL (Í. 36), because AC is equal to CB. (hyp.)

4. Therefore also AL is equal to DF. (ax. 1.)

5. To each of these add CH; therefore the whole AH is equal to DF and CH. (ax. 2.)

6. But AH is contained by AD and DB, since DH is equal to DB. (II. 4, cor.)

7. And DF, together with CH, is the gnomon CMG. 8. Therefore the gnomon CMG is equal to the rectangle AD. DB.

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9. To each of these equals add LG, which is equal to the square on CD. (II. 4, cor., and I. 34.)

10. Therefore the gnomon CMG, together with LG, is equal to the rectangle AD DB, together with the square on CD.

11. But the gnomon CMG and LG make up the whole figure CEFB, which is the square on CB.

12. Therefore the rectangle AD. DB, together with the square on CD, is equal to the square on CB. Conclusion.-Therefore, if a straight line, &c. Q. E. D.

Corollary.-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 and difference.

PROPOSITION 6.-THEOREM.

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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.

(References Prop. I. 31, 34, 36, 43, 46; II. 4, cor.) Hypothesis.—Let the straight line AB be bisected in C, and produced to D.

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

Construction.--1. Upon CD describe the square CEFD (I. 46), and join DE.

2. Through B draw BHG parallel to CE or DF. (I. 31.)

3. Through H draw KLM parallel to AD or EF.

4. And through A draw AK parallel to CL or DM.

Demonstration.--1. Because AC is equal to CB (hyp.), the rectangle AL is equal to CH. (I. 36.)

Ž. But CH is equal to HF (I. 43), therefore AL is equal to HF. (ax. 1.)

3. To each of these add CM; therefore the whole AM is equal to the gnomon CMG. (ax. 2.)

4. But AM is the rectangle contained by AD and DB, since DM is equal to DB. (II. 4, cor.)

5. Therefore the gnomon CMG is equal to the rectangle AD.DB. (ax. 1.)

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6. Add to each of these LG, which is equal to the square on CB. (II. 4, cor., and I. 34.)

7. Therefore the rectangle AD. DB, together with the square on CB, is equal to the gnomon CMG, and the figure LG.

8. But the gnomon ČMG aud LG make up the whole figure CEFD, which is the square on CD.

9. Therefore the rectangle AD DB, together with the square on CB, is equal to the square on CD.

Conclusion. Therefore, if a straight line, &c. Q. E. D.

PROPOSITION 7.-THEOREM.

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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.

(References-Prop. I. 31, 34, 43, 46; II. 4, cor.) Hypothesis.—Let the straight line AB be divided into any two parts in the point C.

Sequence.—The squares on AB and BC shall be equal to twice the rectangle AB. BC, together with the square on AC.

Construction.-1. Upon AB describe the square ADEB (I. 46), and join BD.

2. Through C draw CGF parallel to AD or BE. (I. 31.)

3. Through G draw HGK parallel to AB or DE. (I. 31.)

Demonstration.-1. Then AG is equal to GE. (I. 43.

2. To each of these add CK; therefore the whole AK is equal to the whole CE.

3. Therefore AK and CE are double of AK.

4. But AK and CE are the gnomon AKF, together with the square CK.

5. Therefore the gnomon AKF, together with the square CK, is double of AK.

6. But twice the rectangle AB· BC is also double of AK, for BK is equal to BC. (II. 4, cor.)

7. Therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle AB BC.

8. To each of these equals add HF, which is equal to the square on AC. (II. 4, cor., and I. 34.) 9. Therefore the gnomon AKF, together with the squares

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CK and HF, is equal to twice the rectangle AB· BC, together with the square on AC. 10. But the gnomon AKF, together

with the squares CK and HF, make up the whole figure ADEB and CK, which are the squares on AB and BC.

11. Therefore the squares on AB and BC, are equal to twice the rectangle AB BC, together with the square on AC.

Conclusion.—Therefore, if a straight line, &c. Q. E. D.

PROPOSITION 8.-THEOREM.

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If a straight line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts, together with the square on the other part, is equal to the square on the straight line which is made up of the whole line and the first mentioned part.

(References Prop. I. 3, 31, 34, 36, 43, 46; II. 4. cor.) Hypothesis.—Let the straight line AB be divided into any two parts in the point C.

Sequence.-Four times the rectangle AB BC, together with the square on AC, shall be equal to the square on the straight line made up of AB and BC together.

Construction.–1. Produce AB to D, so that BD may be equal to CB. (post. 2, and I. 3.)

2. Upon AD describe the square AEFD. (I. 46.)

3. And construct two figures such as in the preceding propositions.

Demonstration.-1. Because CB is equal to BD (const.), CB to GK, and BD to KN, (I. 34.)

2. Therefore GK is equal to KN. (ax. 1.)

3. For the same reason PR is equal to RO.

4. And because CB is equal to BD, and GK to KN, 5. Therefore the rectangle CK is equal to BN, and GR to RN. (I. 36.)

6. But CK is equal to RN, because they are complements of the parallelogram Co. (I. 43.).

7. Therefore also BN is equal to GR. (ax. 1.)

8. Therefore the four les BN, CK, GR, RN, are equa to one another, and so the four are quadruple of one of them, CK.

9. Again, because CB is equal to BD, (const.)

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