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It is evident from the course of the proof that the line DF is divided at A so that the rectangle under D F and FA is equal to the on DA. If from A D a part were cut off equal to AG, A D would be divided exactly as AB is divided. But A D is the larger of the two parts into which D F is divided at A, and AF 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 L, ACB. 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 C B by twice the rect. AC, CD.

Since A D is divided into two parts at C, the a on A D is equal to the sum of the Os on A C 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 Don B D 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 Z, therefore the O on AB is equal to the sum of the Os on A D and D B, and the oon C B is equal to the sum of the Os on CD and D B.

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

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 line 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 A in which the _ A B C is an acute

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angle, and let C D be drawn perpendicular to B A, or B A produced.

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

Since AB (in the first figure) and D B (in the second) are divided into two parts, the sum of the Os on A B and B D (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 Os on A B, BD, and C D, is equal to twice the rect. A B, BD, together with the Os on A D and DC.

But since CD B is a right angle, the on CB is equal to the sum of the Os on CD and D B, and the O on C A is equal to the sum of the Os on C D and D A.

Consequently (substituting equals for equals), the sum of the Is on A B and B C is equal to twice the rect. A B, BD, together with the oon CA; that is, the square on C A is less than the sum of the squares on C B 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 Os on one part and the whole is equal to twice the rectangle contained by that part and the whole, together with the o on the other part.”

If the perpendicular drawn from C to the opposite side coincide with one side (CA) of the A, then it is obvious that the con CA is less than the on CB by the 0 on A B (I. 47); consequently, the O on C A is less than the

sum of the Os on A B and B C by twice the on A B, i.e., by twice the rectangle contained by A B and B A.

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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 contained 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.)

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Let A be the given rectilineal figure.* Describe the rectangle BD equal to A. (I. 45.) Produce BC, and cut off C F equal to C D. Bisect BF in 0, and with 0 as centre, at the distance O B or 0 F describe a circle, BGF.

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

Now join O and G.

Because BF is bisected in 0 and divided unequally in C, the rect. B C, CF, together with the on 0 C, is equal to the O on O F, or to the O on O G, which is equal to OF. (II. 5.)

But since O C G is a right-angled A, the on OG is equal to the sum of the Os on 0 C and CG.

Therefore the rect. BC, CF, together with the a on 0 C, is equal to the sum of the Os on 0 C and CG.

Take away the con 0 C from both these equal sums, and it follows that the rect. BC, C F is equal to the O on CG.

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

Therefore the rectangle B D is equal to the a on CG.

But B D is equal to the figure A; therefore the o 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.”

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