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the angle FEG is half a right angle, and therefore equal to the

angle EGF;

wherefore also the side GF is equal to the side FF

And because EC is equal to CA,

the square of EC is equal to the square of CA;

therefore the squares of EC, CA, are double of the square of CA; but the square of EA is equal to the squares of EC, CA; (1. 47) therefore the square of EA is double of the square of AC.

Again, because GF is equal to EF,

the square of GF is equal to the square of EF;

and therefore the squares of GF, FE, are double of the square of EF; but the square of EG is equal to the squares of GF, EF; (1. 47)

therefore the square of EG is double of the square of EF;

and EF is equal to CD; (1. 34)

wherefore the square of EG is double of the square of CD;

But it was demonstrated that the square of EA is 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; (1. 47) therefore the square of AG is double of the squares of AC, CD; but the square of AG is also equal to the squares of AD, DG; (1. 47) therefore the squares of AD, DG, are double of the squares of AC, CD; but 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.-PROBLEM.

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.

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Let AB be the given straight line.

It is required to divide AB into two parts, so that the rectangle

E

contained by the whole and one of the parts shall be equal to the square of the other part.

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Upon AB describe the square ABDC; (1. 46)

bisect AC in E, (1. 10) and join BE;

produce CA to F, and make EF equal to EB; (1. 3)

and upon AF describe the square FGHA. (1. 46.)

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

Produce GH to K.

DEMONSTRATION

Then, because the straight line AC is bisected in E, and produced to the point F,

the rectangle CF, FA, together with the square of AE, is equal to the square of EF; (11. 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 squares of BA, AE, are equal to the square of EB, (1. 47) 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 BA, AE;

take away the square of AE, which is common to both;

therefore the remaining rectangle CF, FA, is equal to the square of AB.

But the figure FK is the rectangle contained by CF, FA, since AF is equal to FG; (def. 30)

and AD is the square of AB;

therefore the figure 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 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. THEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced;

then 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.

(References Prop. 1. 12, 47; II, 4.)

Let ABC be an obtuse-angle triangle, having the obtuse angle ACB; and from the point A let AD be drawn perpendicular to BC produced. (1, 12.)

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

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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; (II. 4)

to each of these equals add the square of DA;

then the squares of BD, DA, are equal to the squares of BC, CD, DA, and twice the rectangles BC, CD;

but the square of BA is equal to the squares of BD, DA, (1. 47) because the angle at D is a right angle;

and the square of CA is equal to the squares of CD, DA; (I. 47)

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, in obtuse-angled triangles, &c.

Q. E. D.

PROP. XIII.-THEOREM.

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

(References

Prop. I. 12, 16, 47; II. 3, 7, 12.)

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. (1. 12.)

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

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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 contained by CB, BD, and the square of DC; (11. 7)

to each of these equals add the square of AD;

therefore the squares of CB, BD, DA, are equal to twice the rectangle CB, BD, and the squares of AD, DC;

but the square of AB is equal to the squares of BD, DA, (1. 47) because the angle BDA is a right angle;

and the square of AC is equal to the squares of AD, DC;

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.

Secondly, Let AD fall without the triangle ABC.

A

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; (11. 12)

to each of these equals add the square of BC,

then the squares of AB, BC, are equal to the square of AC, and twice the square of BC, and twice the rectangle BC, CD;

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; (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 of BC;

therefore the squares of AB, BC, are equal to the square of AC, and twice the rectangle DB, BC;

wherefore 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.

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