The Elements of Euclid, containing the first six books, with a selection of geometrical problems. To which is added the parts of the eleventh and twelfth books which are usually read at the universities. By J. Martin

Construction. Upon AB describe the square ACDB (I. 46); bisect AC in E (I. 10), and join BE; produce CA to F, and make EF equal to EB (I. 3); upon AF describe the square FGĦA (I. 46).

Then AB shall be divided in H, so that the rectangle AB, BH is equal to the square on AH.

Produce GH to meet CD in K.

Proof. Then, because the straight line AC is bisected in E, produced to F, therefore

and

1. The rectangle CF, FA, together with the square on AE, is equal to the square on EF (II. 6);

but EF is equal to EB; therefore

2. The rectangle CF, FA, together with the square on AE, is equal to the square on EB;

but the squares on BA, AE are equal to the square on EB (I. 47), because the angle EAB is a right angle; therefore

3. The rectangle CF, FA, together with the square on AE, is equal to the squares on BA, AE;

take away the square on AE, which is common to both; therefore 4. The rectangle contained by CF, FA is equal to the square on BA.

But the figure FK is the rectangle contained by CF, FA, for FA is equal to FG, and AD is the square on AB; therefore

5. The figure FK is equal to AD;

take away the common part AK, therefore

6. The remainder FH is equal to the remainder HD;

but HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the square on AH; therefore

7. The rectangle AB, BH, is equal to the square on AH. Wherefore the straight line AB is divided in H, so that the rectangle AB, BH is equal to the square on AH. Q.E.F.

PROPOSITION 12.-Theorem.

In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle, is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle.

Let ABC be an obtuse-angled triangle, having the obtuse angle ACB; and from the point A, let AD be drawn perpendicular to BC produced.

Then the square on AB shall be greater than the squares on AC, CB, by twice the rectangle BC, CD.

Demonstration. Because the straight line BD is divided into two parts in the point C, therefore

1. The square on BD is equal to the squares on BC, CD, and twice the rectangle BC, CD (II. 4);

to each of these equals add the square on DA; therefore

2. The squares on BD, DA are equal to the squares on BC, CD, DA, and twice the rectangle BC, CD;

but the square on BA is equal to the squares on BD, DA (I. 47), because the angle at D is a right angle; and the square on CA is equal to the squares on CD, DA; therefore

3. The square on BA is equal to the squares on BC, CA, and

twice the rectangle BC, CD;

that is, the square on BA is greater than the squares on BC, CA, by twice the rectangle BC, CD.

Therefore in obtuse-angled triangles, &c. Q.E.D.

PROPOSITION 13.-Theorem.

In every triangle, the square on the side subtending either of the acute angles, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite angle.

Let ABC be any triangle, and the angle at B one of its acute angles; and upon BC, one of the sides containing it, let fall the perpendicular AD from the opposite angle (I. 12).

Then the square on AC, opposite to the angle B, shall be less than the squares on CB, BA, by twice the rectangle CB, BD.

Demonstration. First. Let AD fall within the triangle ABC. Then, because the straight line CB is divided into two parts in D, 1. The squares on CB, BD are equal to twice the rectangle contained by CB, BD, and the square on DC (II. 7);

to each of these equals add the square on AD; therefore

2. The squares on CB, BD, DA, are equal to twice the rectangle CB, BD, and the squares on AD, DC;

but the square on AB is equal to the squares on BD, DA (I. 47), because the angle BDA is a right angle; and the square on AC is equal to the squares on AD, DC; therefore

3. The squares on CB, BA are equal to the square on AC, and twice the rectangle CB, BD;

that is, the square on AC alone is less than the squares on CB, BA, by twice the rectangle CB, BD.

Secondly. Let AD fall without the triangle ABC.

Then, because the angle at D is a right angle,

1. The angle ACB is greater than a right angle (I. 16);

and therefore

2. The square on AB is equal to the squares on AC, CB, and twice the rectangle BC, CD (II. 12);

to each of these equals add the square on BC; therefore

3. The squares on AB, BC are equal to the square on AC, twice the square on BC, and twice the rectangle BC, CD;

but because BD is divided into two parts in C, therefore

4. The rectangle DB, BC is equal to the rectangle BC, CD, and the square on BC (II. 3);

and the doubles of these are equal; that is, twice the rectangle DB, BC is equal to twice the rectangle BC, CD, and twice the square on BC; therefore

5. The squares on AB, BC are equal to the square on AC, and twice the rectangle DB, BC.

wherefore the square on AC alone is less than the squares on AB, BC; by twice the rectangle DB, BC.

Lastly. Let the side AC be perpendicular to BC.

Then BC is the straight line between the perpendicular and the acute angle at B; and it is manifest, that

The squares on AB, BC, are equal to the square on AC, and twice the square on BC (I. 47).

Therefore in any triangle, &c. Q.E.D.

PROPOSITION 14.-Problem.

To describe a square that shall be equal to a given rectilineal figure. Let A be the given rectilineal figure. It is required to describe a square that shall be equal to A.

Construction. Describe the rectangular parallelogram BCDE equal to the rectilineal figure A (I. 45). Then, if the sides of it, BE, ED, are equal to one another, it is a square, and what was required is now done. But if BE, ED are not equal, produce one of them BE to F, and make EF equal to ED; bisect BF in G (I. 10); from the centre G, at the distance GB or GF, describe the semicircle BHF, and produce DE to meet the circumference in H.

Then the square described upon EH shall be equal to the given rectilineal figure A.

Join GH.

Proof. Then, because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in the point E, therefore

1. The rectangle BE, EF, together with the square on EG, is equal to the square on GF (II. 5);

but GF is equal to GH (Def. 15); therefore

2. The rectangle BE, EF, together with the square on EG, is equal to the square on GH;

but the squares on HE, EG are equal to the square on GH (I. 47); therefore

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The Elements of Euclid, containing the first six books, with a selection of ...