Sidebilder
PDF
ePub

15. And because KF is parallel to HG, and HG parallel to ML, (const.)

16. Therefore KF is parallel to ML. (I. 30.)
17. And KM, FL are parallels, (const.)
18. Therefore KFLM is a parallelogram. (def. 35.)

19. And because the triangle ABD is equal to the parallelogram HF, and the triangle DBC equal to the parallelogram GM, (const.)

20. Therefore the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. (ax. 2.)

Conclusion. Therefore, the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E. Q. E. F.

Corollary. From this it is manifest how to apply to a given straight line a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure; namely, by applying to the given straight line a parallelogram equal to the first triangle ABD, and bav. ing an angle equal to the given angle; and so on. (I. 44.)

PROPOSITION 46.–PROBLEM. To describe a square upon a given straight line. (References, Prop. I. 3, 11, 29, 31, 34; ax. 1, 3; def. 30.) Given. Let AB be the given straight line. Sought.-It is required to describe a square upon AB.

Construction.-1. From the point A draw AC at right angles to AB, (I. 11.)

2. And make AD equal to AB. (I. 3.)

3. Through the point D draw DE C parallel to AB. (1. 31.)

4. Through the point B draw BE parallel to AD. (I. 31.)

Then ADEB shall be the square required.

Proof.-1. Because DE is parallel to AB, and BE parallel to AD (const.), therefore ADEB is a parallelogram.

2. Therefore AB is equal to DE, and AL AD to BE. (I. 34.)

3. But AB is equal to AD. (const.)

4. Therefore the four straight lines BA, AD, DE, EB are equal to one another. (ax. 1.)

5. And the parallelogram ADEB is therefore equilateral.

6. Likewise all its angles are right angles.

7. For since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are together equal to two right angles. (I. 29.)

8. But BAD is a right angle; (const.)
9. Therefore also ADE is a right angle. (ax. 3.)

10. But the opposite angles of parallelograms are equal. (I. 34.)

11. Therefore each of the opposite angles ABE, BED is a right angle. (ax. 1.)

12. Therefore the figure ADEB is rectangular; and it has been proved to be equilateral; therefore it is a square. (def. 30.)

Conclusion.—Therefore, the figure ADEB is a square, and it is described upon the given straight line AB. Q.E.F.

Corollary.-Hence every parallelogram that has one right angle has all its angles right angles.

PROPOSITION 47.—THEOREM. In any right-angled triangle, the square which is described upon the side opposite to the right angle, is equal to the squares described upon the sides which contain the right angle.' (References—Prop. I. 4, 14, 31, 41, 46; ax. 2, 6, 11; def. 30.)

Hypothesis.—Let ABC be a right-angled triangle, having the right-angle BAC.

Sequence. — The square described upon the side BC, shall be equal to the squares described upon BA, AC.

Construction. — 1. On BC describe the square BDEC. (I. 46.)

2. On BA, AC, describe the squares GB, HC. (I. 46.)

3. Through A draw AL parallel to BD or CE (1. 31.)

4. Join AD, FC. Demonstration. — 1. Because the angle BAC is a right angle (hyp.), and that the angle BAG is also a right angle, (def. 30.)

2. The two straight lines AC, AG, upon opposite sides of AB, make with it at the point A, the adjacent angles equal to two right angles;

3. Therefore ĆA is in the same straight line with AG. -(I. 14.)

4. For the same reason, AB and AH are in the same straight line.

5. Now the angle DBC is equal to the angle FBA, for each of them is a right angle (ax. 11), add to each the angle ABC.

6. Therefore the whole angle DBA is equal to the whole angle FBC. (ax. 2.)

7. And because the two sides AB, BD, are equal to the two sides FB, BC, each to each (def. 30), and the angle DBA equal to the angle FBC;

8. Therefore the base AD is equal to the base FC, and the triangle ABD to the triangle FBC. (I. 4.)

9. Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and between the same parallels BD, AL. (1. 41.)

10. And the square GB is double of the triangle FBC, because they are on the same base FB, and between the same parallels FB, GC. (I. 41.)

11. But the doubles of equals are equal (ax. 6), therefore the parallelogram BL is equal to the square GB.

12. In the same manner, by joining AE, BK, it can be shown that the parallelogram CL is equal to the square HC.

13. Therefore the whole square BDEC is equal to the two squares GB, HC. (ax. 2.)

14. And the square BDEC is described on the straight line BC, and the squares GB, HC, upon BA, AC.

15. Therefore the square described upon the side BC, is equal to the squares described upon the sides BA, AC.

Conclusion.—Therefore, in any right-angled triangle, &c. Q. E. D.

PROPOSITION 48.-THEOREM. If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle.

(References—Prop. I. 3, 8, 11, 47; ax. 1, 2.) Hypothesis.—Let the square described upon BC, one of the sides of the triangle ABC, be equal to the squares described upon the other sides BA, AC.

Sequence.—The angle BAC shall be a right angle.

Construction.-1. From the point A draw AD at right angles to AC. (I. 11.)

Ž. Make AD equal to BA (I. 3), and join DC.

Demonstration.-1. Because DA is equal to AB, the square on DA is equal to the square on BA.

2. To each of these add the square on AC.

3. Therefore the squares on DA, AC, are equal to the squares on BA, AC. (ax. 2.)

14. But because the angle DAC is a right angle (const.), the square on DC is equal to the squares on DA, AC. (I. 47.)

5. And the square on BC is equal to B the squares on BA, AC. (hyp.)

6. Therefore the square on DC is equal to the square on BC. (ax. 1.)

7. And therefore the side DC is equal to the side BC.

8. And because the side DA is equal to AB (const.), and AC common to the two triangles DĀC, BAC; the two sides DA, AC, are equal to the two sides BA, AC, each to each;

9. And the base DC has been proved equal to the base BC'; 10. Therefore the angle DAC is equal to BAC. (I. 8.) 11. But DAC is a right angle; (const.) 12. Therefore also BAC is a right angle. (ax. 1.) Conclusion.—Therefore, if the square, &c. Q. E. D.

EXERCISES ON BOOK I.

Prop. 1 to 15. 1. On a given straight line as base, to describe an isosceles triangle having each of its equal sides double of the base.

2. The straight line which bisects the vertical angle of an isosceles triangle, bisects also the base.

3. The straight line drawn from the vertex of an isosceles triangle to the point of bisection of the base, bisects the vertical angle, and is perpendicular to the base.

4. In the figure to Euc. I. 5, if a straight line be drawn from A to the point of intersection of the lines BG, and CF, this line will bisect the angle A.

5. If two triangles on opposite sides of the same base, have the two sides equal which are terminated in one extremity of the base, and likewise those equal which are termi

nated in the other extremity of the base, the two triangles are equal in all respects. To be proved without using I. 8.

6. If two circles cut each other, the straight line which joins their centres bisects that which joins the points of intersection.

7. The straight lines which bisect the three sides of a triangle perpendicularly all meet in the same point.

8. If on a given straight line two isosceles triangles be described (either one on each side, or both on the same side, but not coinciding), the straight line joining their vertices, bisects the given line, and is perpendicular to it.

9. To draw a straight line perpendicular to a given straight line, from its extremity, without producing it.

10. In a given straight line, to find a point equally distant from two given points. When will this be impossible?

11. To describe a circle passing through two given points, and having its centre in a given straight line.

12. If two adjacent angles be bisected, the bisecting lines contain a right angle.

Prop. 16 to 26. 13. Prove that only one perpendicular can be drawn to a given straight line, from a given point, either within or without it.

14. Prove I. 17, without producing any side.

15. Prove I. 18, by cutting off AD=AB, bisecting the angle A, and joining D with the point in which the bisecting line cuts the base.

16. Of all the straight lines which can be drawn to a given straight line from a given point without it, the perpendicular is the least; of the others, one nearer to the least is less than one more remote; and from the same point only two equal straight lines can be drawn, one upon each side of the perpendicular.

17. Prove I. 20, without producing any side, by bisecting one of the angles.

18. The difference of any two sides of a triangle is less than the third side.

19. If from any point within a triangle straight lines be drawn to the vertices of the three angles, these three straight lines are less than the perimeter of the triangle, but greater than half the perimeter.

20. To describe a rectilineal figure equal in all respects to a given rectilineal figure.

21. Given two sides of a triangle, and one of the angles, to construct the triangle. When will there be two solutions, when one, and when none?

« ForrigeFortsett »