Imágenes de páginas
PDF
EPUB

vertically opposite angle EBC,) and that BDG is a right angle, (for it is equal to the alternate angle DCE,) therefore the remaining angle DGB is half a right angle, and equal to the angle DBG, and therefore also the side DG is equal to the side DB: Again, because FGE is half a right angle, and EFG a right angle, (for it is equal to the opposite angle ECD,) therefore the remaining angle FEG is half a right angle, and equal to

E

F

B

G

the angle FGE, and therefore also the side FE is equal to the side FG.

Now, because AC is equal to CE, the square of AC is equal to the square of CE, and therefore the squares of AC, CE are double of the square of AC: But the square of AE is equal to the squares of AC, CE; therefore the square of AE is double of the square of AC: Again, because EF is equal to FG, the square of EF is equal to the square of FG, and therefore the squares of EF, FG are double of the square of EF: But the square of EG is equal to the squares of EF, FG; therefore the square of EG is double of the square of EF, that is, of the square of CD: And the square of AE is also double of the square of AC; therefore the squares of AE, EG are double of the squares of AC, CD: But the square of AG is equal to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD: And the square of AG is equal to the squares of AD, DG; therefore the squares of AD, DG are double of the squares of AC, CD: And DG is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD.

Wherefore, If a straight line &c.

PROP. XI. PROB.

To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part.

Let AB be the given straight line: it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part.

Upon AB describe the square ACDB; bisect AC in E, and join BE; produce CA to F, and make EF equal to EB; and upon AF describe the square AFGH: AB is divided in H, so that the rectangle AB, BH is equal to the square of AH.

F

A

E

G

H

B

C K D

Produce GH to K: Then, because the straight line AC is bisected in E, and produced to F, the rectangle CF, FA, together with the square of AE, is equal to the square of EF (2. 6): But EF is equal to EB; therefore the rectangle CF, FA, together with the square of AE, is equal to the square of EB: And the square of EB is equal to the squares of AE, AB, because the angle EAB is a right angle; therefore the rectangle CF, FA, together with the square of AE, is equal to the squares of AE, AB: Take away the common square of AE; therefore the remainder, the rectangle CF, FA, is equal to the square of AB: And the figure FK is the rectangle contained by CF, FA, for FG is equal to FA, and AD is the square of AB; therefore FK is equal to AD: Take away the common part AK, and 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 of AH; therefore the rectangle AB, BH is equal to the square of AH: Wherefore the given straight line AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. Q. E. F.

PROP. XII. THEOR.

In obtuse-angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of 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: the square of AB is greater than the squares of AC, CB, by twice the rectangle BC, CD.

B O

Α

Because the straight line BD is divided into two parts in the point C, the square of BD is equal to the squares of BC, CD, and twice the rectangle BC, CD (2.4): To each of these equals add the square of DA; therefore the squares of BD, DA are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD: But the square of BA is equal to the squares of BD, DA, because the angle at D is a right angle, and the square of CA is equal to the squares of CD, DA; therefore the square of BA is equal to the squares of BC, CA, and twice the rectangle BC, CD, that is, the square of BA is greater

than the squares of BC, CA, by twice the rectangle BC, CD.

Wherefore, If obtuse-angled triangles &c. Q. E. D.

PROP. XIII. THEOR.

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

D

First, let AD fall within the triangle ABC: Then, because the straight line CB is divided into two parts in the point D, the squares of CB, BD are equal to twice the rectangle CB, BD, and the square of CD (2.7): To each of these equals add the square of DA; therefore the squares of CB, BD, DA are equal to twice the rectangle CB, BD, and the squares of CD, DA: But the square of AB is equal to the squares of BD, DA, (because the angle BDA is a right angle,) and the square of AC is equal to the squares of CD, DA; therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD, that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB, BD.

B

A

Next, let AD fall without the triangle ABC: Then, because the angle at D is a right angle, the angle ACB is greater than a right angle (1. 16); and therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle BC, CD (2. 12): To each of these equals add the square of BC; therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle BC, CD, and twice the square of BC: But because BD is divided into two parts in C, the rectangle DB, BC is equal to the rectangle BC, CD, and the square of BC (2. 3); therefore twice the rectangle DB, BC is equal to twice the rectangle BC, CD, and twice the square of BC; and therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC, that is, the square of AC alone is less than the squares of 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 of AB, BC, are equal to the square of AC and twice the square of BC (1. 47).

Wherefore, In every triangle &c. Q.E.D.

PROP. XIV. PROB.

A

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.

Describe (1. 45) a rectangular parallelogram BCDE equal to the rectilineal figure A: If then the sides

« AnteriorContinuar »