Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

ACB, and from the point A let AD be drawn (12. 1.) perpendicular to BC produced: The square of AB is greater than the squares of AC, CB, by twice the rectangle BC.CD.

=

A

Because the straight line BD is divided into two parts in the point C, BD2 (4. 2.) BC + CD3 + 2BC.CD; add AD to both; Then BD2 + AD2 = BC2 + CD3+AD3 +2BC.CD. But AB = BD2 + AD (47. 1.), and AC-CD+ AD2 (4. 1.); therefore, ABBC+AC2 + 2BC.CD; that is, AB is greater than BC+AC2 by 2BC.CD. Therefore, in obtuse angled triangles, &c. Q. E. D.

B

PROP. XIII, THEOR.

C

D

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 (12. 1.) 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.

First let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D (7. 2.), BC+BD3= 2BC.BD+CD2. Add to each AD3; then BC BD + AD2 : 2BC.BD +CD2+AD3. But BD+AD2 AB, and CD3 + DA3 (47. 1.); therefore BC+ AB3 =

=

[ocr errors]

AC

2BC.BD+AC2; that is, AC is less B than BC+AB by 2BC.BD.,

A

D

+ Secondly, Let AD fall without the triangle ABC*: Then because the angle at D is a right angle, the angle ACB is greater (16. 1.) than a right angle, and AB (12. 2.) AC+BC2+2BC.CD. Add BC2 to each; then AB+BC-AC+2BC+2BC.CD. But because BD is divided into two parts in C, BC+BC.CD= (3. 2.) BC.BD, and 2BC+2BC.CD=2BC.BD: therefore AB + BC3=A€3+ 2BC.BD; and AC is less than AB2+BC3, by 2BD.BC.

* See figure of the last Proposition.

Lastly, Let the side AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifest that (47. 1.) AB+BC2=AC®+2BC2=AC2 +2BC.BC. Therefore in every triangle, &c. Q. E. D.

PROP. XIV. PROB.

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 (45.1.) the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it, BE, ED are equal to one another, it is a square, and what was required is done; but if they are not equal, produce one of them, BE to F, and make EF equal to ED, and bisect BF in G: and from the centre G, at the distance GB, or GF, describe the semicircle BHF and produce DE to H, and join GH. Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the rectangle BE.EF, together with the square of EG, is equal (5. 2.) to the square of GF: but GF is equal to GH; therefore the rectangle BE.EF, together with the square of EG, is equal to the square of GH: But the squares of HE and EG are equal (47. 1.) to the square of GH: Therefore also the rectangle BE.EF together with the square of EG,is equal to the squares of HE and EG. Take

away the square of EG, which is common to both, and the remaining rectangle BE.EF is equal to the square of EH: But

BD is the rectangle con

H

[blocks in formation]

tained by BE and EF, because FF is equal to ED; therefore BD is equal to the square of EH; and BD is also equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. Which was to be done.

[ocr errors][merged small]

If one side of a triangle be bisected, the sum of the squares of the other two sides is double of the square of half the side bisected, and of the square of the line drawn from the point of bisection to the opposite angle of the triangle.

Let ABC be a triangle, of which the side BC is bisected in D, and DA drawn to the opposite angle; the squares of BA and AC are together double of the squares of BD and DA.

From A draw AE perpendicular to BC, and because BEA is a right angle, AB (47. 1.) BE+AE and

=

AC2 = CE2 + AE; wherefore AB2+
AC BE+ CE2 + 2AE2. But be-
cause the line BC is cut equally in D,
and unequally in E, BE+ ČE2
(9. 2.) 2BD+2DE2; therefore AB2+
AC2=2BD2+2DE2. 2AE2.

Now DE2+ AE2 (47. 1.) AD2, and 2DE2 + 2AE2 = 2AD2; wherefore AB2+AC2-2BD2+2AD2, There- B fore, &c. Q. E. D.

PROP. B. THEOR.

A

[blocks in formation]

The sum of the squares of the diameters of any parallelo gram is equal to the sum of the squares of the sides of the parallelogram.

Let ABCD be a parallelogram, of which the diameters are AC and BD; the sum of the squares of AC and BD is equal to the sum of the squares of AB, BC, CN, DA.

A

E

D

Let AC and BD intersect one another in E: and because the vertical angles AED, CEB are equal (15. 1.) and also the alternate angles EAD, ECB (29. 1.), the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each; but the sides AD and BC, which are opposite to equal angles in these triangles, are also equal (34. 1.); therefore the other sides which are opposite to the equal angles are also equal (26. 1.), viz. AE to EC, and ED to EB.

B

Since, therefore, BD is bisected in E, AB3+ADo=(A. 2.) 2BE2 +2AE2; and for the same reason, CD3+BC=2BE2+2EC=2BE* +2AE, because EC-AE. Therefore AB+AD2+DC3+BC2= 4BE+4AE. But 4BE-BD, and 4AE-AC (2. Cor. 8. 2.) because BD and AC are both bisected in E; therefore AB+AD2+ CD2+BC=BD+AC2. Therefore the sum of the squares &c. Q.E.D. COR. From this demonstration, it is manifest that the diameters of every parallelogram bisect one another.

ELEMENTS

OF

GEOMETRY.

BOOK III.

DEFINITIONS.

A.

HE radius of a circle is the straight line drawn from the centre to the circumference.

THE

[blocks in formation]

An arch of a circle is any part of the circumference.

V.

A segment of a circle is the figure contained by a straight line, and the arch which it cuts off.

I

VI.

An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment.

VII.

And an angle is said to insist or stand upon the arch intercepted between the straight lines which contain the angle.

VIII.

The sector of a circle is the figure contained by two straight lines drawn from the centre, and the arch of the circumference between them.

IX.

Similar segments of a circle, are those in which the angles are equal, or which contain equal angles.

PROP. I. PROB.

To find the centre of a given circle.

Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisect (10. 1.) it in D; from the point D draw (11. 1.) DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the cir

ele ABC.

For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are radii of the same circle: therefore the angle ADG is equal (8. 1.) to the angle GDB: But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle (7. def. 1.) Therefore the angle GDB is a right angle: But FDB is likewise a right angle; wherefore the angle FDB is equal to the angle GDB, the greater

A

F

G

B

D

E

« ΠροηγούμενηΣυνέχεια »