Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

It is evident from the ccurse of the proof that the line DF is divided at A so that the rectangle under DF and FA is equal to the on DA. If from AD a part were cut off equal to A G, A D would be divided exactly as A B is divided. But AD is the larger of the two parts into which D F is divided at A, and A F is the smaller. It follows, therefore, that if a line be divided in extreme and mean ratio, and the smaller segment be measured off upon the larger, this larger segment will itself be divided in a similar manner. It is clear, also, that this process admits of being continued ad infinitum.

PROPOSITION XII.

In an obtuse-angled triangle the square on the side subtending the obtuse angle exceeds the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by either of them and the distance between a perpendicular drawn to that side produced from the opposite angle, and the point of the obtuse angle.

For the proof, we must know, —

1. That in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

2. That if a line be divided into any two parts, the square on the whole line is equal to the sum of the squares on the parts, together with twice the rectangle contained by the parts. (II. 4.)

Let A B C be a ▲ having an obtuse, A CB. Let BD be drawn perpendicular to AC E produced.

We have to prove that the on A B exceeds the sum of the Os on AC and CB by twice the rect. A C, CD.

Since A D is divided into two parts at C, the on AD is equal to the sum of the Os on AC and CD, together with twice the rect. A C, CD. (II. 4.)

To each of these equals add the square on B D. Then the on A D together with the on BD is equal to the sum of the Os on A C, CD, and BD, together with twice the rect. A C, C D.

But the

at D is a right, therefore the on A B is equal to the sum of the s on A D and DB, and the on CB is equal to the and D B.

sum of the Os on CD

on A B is equal to

It follows, therefore, that the the sum of the Os on AC and CB, together with twice the rect. A C, CD; that is, the on AB exceeds the sum of the s on AC and CB by twice the rect. A C, CD.

PROPOSITION XIII.

In any triangle the square on the side subtending an acute angle is less than the sum of the squares on the sides containing that angle by twice the rectangle contained by either of them and the distance between a linə drawn perpendicular to this side (or its production) from the opposite angle, and the point of the acute angle.

For the proof of this proposition we must know,— 1. That in a right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides.

2. That if a line be divided into any two parts, the sum of the squares on the whole line and one part is equal to twice the rectangle contained by the whole and that part, together with the square on the other part. (II. 7.)

Let ABC be a ▲ in which the ABC is an acute

ДА

A

D

B D

A

B

angle, and let CD be drawn perpendicular to BA, or BA produced.

We have to prove that the☐ on AC is less than the sum of the s on A B and B C by twice the rect. A B, BD.

Since AB (in the first figure) and DB (in the second) are divided into two parts, the sum of the s on AB and BD (in either case) is equal to twice the rect. A B, BD, together with the □ on A D.* (II. 7.)

To each of these equals add the on CD; then the sum of the s on A B, BD, and CD, is equal to twice the rect. A B, BD, together with the □s on AD and D C.

But since CDB is a right angle, the on CB is equal to the sum of the Os on CD and D B, and the on CA is equal to the sum of the Os on CD and DA.

Consequently (substituting equals for equals), the sum of the Os on A B and BC is equal to twice the rect. A B, BD, together with the on CA; that is, the square on CA is less than the sum of the squares on CB and A B by twice the rect. A B, B D.

* The beginner will notice that in applying Prop. 7 to the second figure it must be read thus:-"If a line be divided into two parts, the sum of the s on one part and the whole is equal to twice the rectangle contained by that part and the whole, together with the on the other part."

A

B

the

If the perpendicular drawn from C to the opposite side coincide with one side (CA) of the A, then it is obvious that the on CA is less than the on

CB by the

quently, the

on AB (I. 47); conse

on CA is less than the

sum of thes on A B and B C by twice

on AB, i.e., by twice the rectangle contained

by A B and B A.

PROPOSITION XIV.

To describe a square that shall be equal to a given rectilineal figure.

For the construction we must be able,

1. To produce a given finite right line. (Post. 2.)
2. To describe a circle with a given point or
centre, and at a given distance from that centre.
(Post. III.)

3. From the greater of two given right lines to
cut off a part equal to the less. (I. 3.)
4. To describe a rectangular parallelogram equal
to a given rectilineal figure. (I. 45.)

5. To bisect a given finite right line. (I. 10.)
For the proof we must know (besides the axioms),—
1. That in a right-angled triangle the square
on the hypotenuse is equal to the sum of the
squares on the other two sides. (I. 47.)
2. That if a line be divided into two equal, and
also into two unequal parts, the rectangle con-
tained by the unequal parts together with the
square on the part between the points of section
is equal to the square on half the line. (II. 5.)

Let A be the given rectilineal figure.* Describe the

rectangle BD equal to A. (I. 45.) Produce BC, and cut off CF equal to CD. Bisect B F

in O, and with O

as centre, at the

B

E

distance O B or O F describe a circle, B G F.

[merged small][ocr errors]

Produce DC to meet the circle in G. The on CG will be the

required.

Now join O and G.

Because BF is bisected in O and divided unequally

in C, the rect. BC, CF, together with the

is equal to the

is equal to O F.

on O C,

on OF, or to the on O G, which (II. 5.) But since O C G is a right-angled ▲, the on O G is equal to the sum of the Os on OC and CG.

Therefore the rect. BC, CF, together with the on O C, is equal to the sum of the s on OC and CG. Take away the on OC from both these equal sums, and it follows that the rect. BC, C F is equal to the on C G.

But the rect. B C, C F is the same as the rect. BD, for CF is equal to C D.

Therefore the rectangle BD is equal to the on

CG.

But BD is equal to the figure A; therefore the ☐ on CG is equal to the figure A.

*The introduction of this rectilineal figure is a perfectly needless proceeding. The proposition is (essentially), “ To construct a square equal to a given rectangle."

« ΠροηγούμενηΣυνέχεια »