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Again, because the angle at B is half a right angle,
and FDB a right angle,

for it is equal to the interior and opposite angle ECB, (1. 29.)
therefore the remaining angle BFD is half a right angle;
wherefore the angle at B is equal to the angle BFD,
and the side DF equal to the side DB. (1. 6.)
And because AC is equal to CE,

the square on AC is equal to the square on CE;

therefore the squares on AC, CE are double of the square on AC; but the square on AE is equal to the squares on AC, CE, (1. 47.) because ACE is a right angle;

therefore the square on AE is double of the square on AC.
Again, because EG is equal to GF,

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

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

therefore the square on EF is double of the square on CD; but the square on AE is double of the square on AC; therefore the squares on AE, EF are double of the squares on AC, CD; but the square on AF is equal to the squares on AE, EF, because AEF is a right angle: (1.47.)

therefore the square on AF is double of the squares on AC, CD: but the squares on AD, DF are equal to the square on AF; because the angle ADF is a right angle; (1. 47.)

therefore the squares on AD, DF are double of the squares on 4C, CD; and DF is equal to DB;

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

PROPOSITION X. THEOREM.

If a straight line be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D.

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Then the squares on AD, DB, shall be double of the squares on AC. CD.

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From the point C draw CE at right angles to AB, (1. 11.).
make CE equal to AC or CB, (1. 3.) and join AE, EB;
through E draw EF parallel to AB, (I. 31.)
and through D draw DF parallel to CE, meeting EF in F.

Then because the straight line EF meets the parallels CË, 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, (I. 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 Fis 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; but 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 ÊA, 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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H

B

K D

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, (1. 3.)
upon AF describe the square FGHA. (1.46.)

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

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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.)

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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, (1. 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 ID;
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, EH, 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,

because the angle at D is a right angle;

(1. 47.) ́

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, and twice the rectangle BC, CD;

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.)

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Then the square on AC opposite to the angle B, shall be less thna 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, BĂ, 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, C 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 A3, 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 DR, 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 Calone 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.

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