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to the triangle EBC, because it is upon the same base BC, and between the same parallels

Α.

D BC, AE ; But the triangle ABC is equal to the triangle BDC ; therefore also the triangle BDC is equal to the triangle EBC, the greater to the less, which is impossible: Therefore AE is not parallel to BC. In the same manner, it may be demonstrated that no other line but AD is parallel to BC; AD is therefore parallel to it.

B PROP. XXXIV. THEOR. Equivalent triangles on the same side of bases which are equal and in the

same straight line, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and towards the same parts ; they are be

A

D tween the same parallels.

Join AD; AD is parallel to BC; For, if it is not, through A draw AG parallel to BF, and join GF. The triangle ABC is equal (32. 1.) to the triangle GEF, because they are B upon equal bases BC, EF, and between the same parallels BF, AG; But the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible ; Therefore AG is not parallel to BF; and in the same manner it may be demonstrated that there is no other parallel to it but AD; AD is therefore parallel to BF.

PROP. XXXV. THEOR. If a parallelogram and a triangle be upon the same base, and between the

same parallels ; the parallelogram is double of the triangle. Let the parallelogram ABCD and the triangle EBC be upon the same base BC and

A

D between the same parallels BC, AE; the parallelogram ABCD is double of the triangle EBC.

Join AC; then the triangle ABC is equal (31. 1.) to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram ABCD is double (28. 1.) of the trian B

Cai gle ABC, because the diameter AC divides EBC.

th

PROP. XXXVI. THEOR.

The complements of the parallelograms which are about the diameter of any

parallelogram, are equivalent to one another. Let ABCD be a parallelogram of which the diameter is AC; let EH, FG be the parallelograms about AC, that is, through which AC passes, and let BK, KD be the other parallelograms, which make up the whole figure A H

D ABCD, and are therefore called the complements; The complement BK is equal

K to the complement KD.

E Because ABCD is a parallelogram and AC its diameter, the triangle ABC is equal (28. 1.) to the triangle ADC; And because EKHA is a parallelogram, and AK its diameter, the triangle AEK

B G is equal to the triangle AHK : For the same reason, the triangle KGC is equal to the triangle KFC. Then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to the triangle KFC; the triangle AEK, together with the triangle KGC, is equal to the triangle AHK, together with the triangle KFC: But the whole triangle ABC is equal to the whole ADC ; therefore the remaining complement BK is equal to the remaining complement KD.

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In any right angled triangle, the square which is described upon the side sub

tending the right angle, is equivalent to the squares described upon the sides which contain the right angle.

Let ABC be a right angled triangle having the right angle BAC;

G the square described upon

the side BC is equal to the squares described

H upon BA, AC.

On BC'describe the square BDEC, and on BA, AC the squares GB, HC;

K and through A draw AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC,

B

ok BAG is a right angle (24. def.), the two straight lines AC, AG upon the opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles ; therefore CA is in the same straight line (7. 1.) with AG; for the same reason, AB

LEO and AH are in the same straight line.

Now because the angle DBC is equal to the angle FBA, each of them being a right angle, adding to each the angle ABC, the whole angle DBA will be equal (2. Ax.) to the whole FBC; and because the two sides AB, BD, are equal to the two FB, BC each to each, and the angle DBA equal to the angle FBC, therefore the base AD is equal (1. 1.) to the base FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double (35. 1.) of the triangle ABD, because they are upon the same base, BD, and between the same parallels, BD, AL; and the square GB is double of the triangle BFC because these also are upon the same base FB, and between the same parallels FB, GC. Now the doubles of equals are equal (6. Ax.) to one another; therefore the parallelogram BL is equal to the square GB : And in the same manner, by joining AE, BK, it is demonstrated, that the parallelogram CL is equal to the square HC. Therefore, the whole square BDEČ is equal to the two squares GB, HC; and the square

BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC: wherefore the square upon the side BC is equal to the squares upon the sides BA, AC.

Cor. 1. Hence, the square of one of the sides of a right angled triangle is equivalent to the square of the hypotenuse diminished by the square of the other side ; which is thus expressed: AB2=BC2-AC2.

Cor. 2. If AB=AC; that is, if the triangle ABC be right angled and isosceles ; BC2=2AB2=2AC2; therefore, BC=AB V2.

Cor. 3. Hence, also, if two right angled triangles have two sides of the one, equal to two corresponding sides of the other.; their third sides will also be equal, and the triangles will be identical.

PROP. XXXVIII. THEOR.

If the square described upon one of the sides of a triangle, be equivalent to the

squares described upon the other two sides of it; the angle contained by these two sides is a right angle.

If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle.

From the point A draw AD at right angles to AC, and make AD equal to BA, and join DC. Then because DA is equal to AB, the square of DA is equal to the square of

DA AB; to each of these add the

square of AC; therefore the squares of DA, AC are equal to the squares of BA, AC. But the square of DC is equal (37. 1.) to the squares of DA, AC, because DAC is a right angle; and the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore, the square of DC is equal to the square of BC; and therefore also the side DC is equal to B the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, and the base DC likewise equal to the base BC, the angle DAC is equal (5. 1.) to the angle BAC; But DAC is a right angle ; therefore also BAC is a right angle.

PROBLEMS

RELATING TO THE FIRST BOOK.

PROP. I. PROBLEM.

To describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it. From the centre A, at the dis

C tance AB, describe (3. Postulate) the circle BCD, and from the cen- , tre B, at the distance BA, describe the circle ACE ; and from the point

D. A C, in which the circles cut one

B

E another, draw the straight lines (1. Post.) CA, CB to the points A, B; ABC is an equilateral triangle.

Because the point A is the centre of the circle BCD, AC is equal (11. Def.) to AB; and because the point B is the centre of the circle ACE, BC is equal to AB: But it has been proved that CA is equal to AB; therefore CA, CB are each of them equal to AB; now things which are equal to the same are equal to one another (1. Ax.); therefore CA is equal to CB; wherefore, CA, AB, CB are equal to one another; and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB.

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PROP. II. PROB.

From a given point to draw a straight line equal to a given straight line.

K

H

Let A be the given point, and BC the given straight line; it is required to draw, from the point A, a straight line equal to BC.

From the point to B draw (1. Post.) the straight line AB; and upon it describe (Prob. 1.) the equilateral triangle DAB, and produce (2. Post.) the straight lines DA, DB, to E and F; froin the centre B, at the distance BC, describe (3. Post.) the circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL is equal to BC.

Because the point B is the centre of the circle CGH, BC is equal (11. Def.) to BG; and because D is the centre of the circle GKL,

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DL is equal to DG, and DA, DB, parts of them, are equal ; therefore the remainder AL is equal to the remainder (3. Ax.) BG: But it has been shewn that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line AL is equal to BC. Wherefore, from the given point A, a straight line AL has been drawn equal to the given straight line BC.

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From the greater of two given straight lines to cut off a part equal to the less

Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less.

From the point A draw (Prob. 2.) the straight line AD equal to C ; and from the

А

E B centre A, and at the distance AD, describe (3. Post.) the circle DEF; and because A is the centre of the circle DEF, AE is equal to AD; but the straight line

F C is likewise equal to AD; whence AE and C are each of them equal to AD; wherefore the straight line AE is equal to (1. Ax.) C, and from AB the greater of two straight lines, a part AE has been cut off equal to C the less.

PROP. IV. PROB.

To bisect a given rectilineal angle, that is, to divide it into two equal angles.
Let BAC be the given rectilineal angle, it is required to bisect it.

Take any point D in AB, and from AC cut (Prob. 3.) off AE equal to AD; join DE, and upon it describe (Prob. 1.) an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC.

10: Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal to the two D sides EA, AF, each to each ; but the base DF is also equal to the base EF; therefore the angle DAF is equal (5. 1.) to the angle EAF : wherefore the given rectilineal angle BAC is bisected by the straight line

F AF.

B

SCHOLIUM. By the same construction, each of the halves BAF, CAF, may be divided into two equal parts ; and thus, by successive subdivisions, a given angle may be divided into four equal parts, into eight, into sixteen, and so on.

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