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Then because the straight line EF meets the parallels CE, FD, therefore the angles CEF, EFD are equal to two right angles; (1. 29.; and therefore the angles BEF, EFD are less than two right angles.

But straight lines, which with another straight line make the interior angles upon the same side of a line, less than two right angles, will meet if produced far enough; (I. ax. 12.)

therefore EB, FD will meet, if produced towards B, D;
let them be produced and meet in G, and join AG.
Then, because AC is equal to CE,

therefore the angle CEA is equal to the angle EAC; (1. 5.)
and the angle ACE is a right angle;

therefore each of the angles CEA, EAC is half a right angle. (1. 32.) For the same reason,

each of the angles CEB, EBC is half a right angle;
therefore the whole AEB is a right angle.
And because EBC is half a right angle,
therefore DBG is also half a right angle, (1. 15.)
for they are vertically opposite;

but BDG is a right angle,

because it is equal to the alternate angle DCE; (1. 29.)
therefore the remaining angle DGB is half a right angle;
and is therefore equal to the angle DBG;

wherefore also the side BD is equal to the side DG. (1. 6.) Again, because EGF is half a right angle, and the angle at F is a right angle, being equal to the opposite angle ECD, (1. 34.) therefore the remaining angle FEG is half a right angle, and therefore equal to the angle EGF;

wherefore also the side GF is equal to the side FE. (1.6.)
And because EC is equal to CA;

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

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

the square on GF is equal to the square on FE;

therefore the squares on GF, FE are double of the square on FE; put the square on EG is equal to the squares on GF, FE; (1. 47.) therefore the square on EG is double of the square on FE; and FE is equal to CD; (1. 34.)

wherefore the square on EG is double of the square on CD; but it was demonstrated,

that the square on EA is double of the square on AC; therefore the squares on EA, EG are double of the squares on AC, CD; but the square on AG is equal to the squares on EA, EG; (1. 47.) therefore the square on AG is double of the squares on AC, CD: but the squares on AD, DG are equal to the square on AG; therefore the squares on AD, DG are double of the squares on AC, CD; but DG is equal to DB;

therefore the squares on AD, DB are double of the squares on AC, CD. Wherefore, if a straight line, &c. Q. E.D.

PROPOSITION 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 on the other part.

Let AB be the given straight line.

It is required to divide AB into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square on the other part.

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Upon AB describe the square ACDB; (1. 46.)
bisect AC in E, (1. 10.) and join BE,

produce CA to F, and make EF equal to EB, (T. 3.)
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 on AH.

Produce GH to meet CD in K.

Then because the straight line AC'is bisected in E, and produced to F, therefore the rectangle CF, FA together with the square on AE, is equal to the square on EF; (11. 6.)

but EF is equal to EB;

therefore the rectangle CF, FA together with the square on AE, is equal to the square on EB;

but the squares on BA, AE are equal to the square on EB, (I. 47.) because the angle EAB is a right angle;

therefore the rectangle CF, FA, together with the square on AE, is equal to the squares on BA, AE;

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

therefore the rectangle contained by CF, FA is equal to the square on BA.

But the figure FK is the rectangle contained by CF, FA,
for FA is equal to FG;

and AD is the square on AB;
therefore the figure FK is equal to AD;
take away the cominon part AK,

therefore 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 on AH;

therefore the rectangle AB, BH, is equal to the square on AH. Wherefore the straight line AB is divided in H, so that the rectangle AB, BH is equal to the square on AH. Q. E. F.

PROPOSITION XII. THEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle, is greater than the squares on 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.

Then the square on AB shall be greater than the squares on 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, therefore the square on BD is equal to the squares on BC, CD, and twice the rectangle BC, CD; (II. 4.)

to each of these equals add the square on DA;

therefore the squares on BD, DA are equal to the squares on BC, CD, DA, and twice the rectangle BC, CD;

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

and the square on CA is equal to the squares on CD, DA; therefore the square on BA is equal to the squares on BC, CA, twice the rectangle BC, CD;

and

that is, the square on BA is greater than the squares on BC, CA, by twice the rectangle BC, CD.

Therefore in obtuse-angled triangles, &c. Q.E.D.

PROPOSITION XIII. THEOREM.

In every triangle, the square on the side subtending either of the acute angles, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite 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. (1. 12.)

Then the square on AC opposite to the angle B, shall be less than the squares on 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 D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD, and the square on DC; (II. 7.)

to each of these equals add the square on AD;

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

but the square on AB is equal to the squares on BD, DA, (1.47.) because the angle BDA is a right angle;

and the square on AC is equal to the squares on AD, DC; therefore the squares on CB, BA are equal to the square on AC, and twice the rectangle CB, BD:

that is, the square on AC alone is less than the squares on CB, BA, by twice the rectangle CB, BD. Secondly, let AD fall without the triangle ABC.

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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 on AB is equal to the squares on AC, CB. and twice the rectangle BC, CD; (II. 12.)

to each of these equals add the square on BC;

therefore the squares on AB, BC are equal to the square on AC, twice the square on BC, and twice the rectangle BC, CD; but because BD is divided into two parts in C,

therefore the rectangle DB, BC is equal to the rectangle BC, CD, and the square on 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 on BC:

therefore the squares on AB, BC are equal to the square on AC, and twice the rectangle DB, BC:

wherefore the square on AC alone is less than the squares on AB, BC; by twice the rectangle DB, BC.

Lastly, let the side AC be perpendicular to BC.

A

B

Then BC is the straight line between the perpendicular and the acute angle at B;

and it is manifest, that the squares on AB, BC, are equal to the square on AC, and twice the square on BC. (1.47.)

Therefore in any triangle, &c. Q. E.D.

F

PROPOSITION XIV. PROBLEM.

To describe a square that shall be equal to a given rectilineai figure.
Let A be the given rectilineal figure.

It is required to describe a square that shall be equal to A.

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Describe the rectangular parallelogram BCDE equal to the rectilineal figure A. (1.45.)

Then, if the sides of it, BE, ED, are equal to one another,
it is a square, and what was required is now done.

But if BE, ED, are not equal,

produce one of them BE to F, and make EF equal to ED,
bisect BF in G; (1. 10.)

from the center G, at the distance GB, or GF, describe the semicircle BHF,

and produce DE to meet the circumference in H.

The square described upon EH shall be equal to the given rectilineal figure A.

Join GH.

Then because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in the point E;

therefore the rectangle BE, EF, together with the square on EG, is equal to the square on GF; (11. 5.)

but GF is equal to GH; (def. 15.)

therefore the rectangle BE, EF, together with the square on EG, is equal to the square on GH;

but the squares on HE, EG are equal to the square on GH; (1. 47.) therefore the rectangle BE, EF, together with the square on EG, is equal to the squares on HE, EG;

take away the square on EG, which is common to both; therefore the rectangle BE, EF is equal to the square on HE. But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED;

therefore BD is equal to the square on EH;

but BD is equal to the rectilineal figure A; (constr.)

therefore the square on EH is equal to the rectilineal figure A. Wherefore a square has been made equal to the given rectilineal figure A, namely, the square described upon EH. Q.E.F.

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