therefore the remaining angle EFG is half a right angle;
wherefore the angle GEF is equal to the angle EFG,
and the side EG equal to the side GF. (1. 6.)

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 of AC is equal to the square of CE;

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

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

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

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

therefore the square of EF is double of the square of CD ;
but the square of AE is double of the square of AC;

therefore the squares of AE, EF are double of the squares of AC, CD; but the square of AF is equal to the squares of AE, EF, because AEF is a right angle: (1. 47.)

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

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

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

PROPOSITION X. THEOREM.

Q. E. D.

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

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

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, (1. 31.)
and through D draw DF parallel to CE.

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 less than two right angles, will meet if duced far enough; (ax. 12.)

therefore EB, FD will meet, if produced towards B, D ;
let them 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;

pro

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 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;

therefore the squares of GF, FÊ 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 EA, EG are double of the squares of AC, CD; but the square of AG is equal to the squares of EÂ, EG; (1. 47.) therefore the square of A G is double of the squares of AC, CD: but the squares of AD, DG are equal to the square of AG; 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.

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 of 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 and one of the parts, shall be equal to the square of the other part.

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.)

Then AB shall be divided in H, so that the rectangle AB, BH is equal to the square of 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 of AE, is equal to the square of EF; (II. 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;

but 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, AË;

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

therefore the rectangle contained by CF, FA is equal to the square of BA. But the figure FK is the rectangle contained by CF, FA, for FA is equal to FG;

and AD is the square of AB; therefore the figure FK is equal to AD;

take away the common part A K,

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 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.

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 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.

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

A

B

Because the straight line BD is divided into two parts in the point C, therefore 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 ᎠᎪ ;

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

DA;

and the square of CA is equal to the squares of CD, 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.

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

PROPOSITION 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 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 of AC opposite to the angle B, shall be less than the squares of CB, BA, by twice the rectangle CB, DB.

First, let AD fall within the triangle ABC.

Then, because the straight line CB is divided into two parts in 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.

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;

therefore the squares of AB, BC are equal to the square of AC, twice the square of 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 of BC; (11. 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:

wherefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC:

therefore 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.

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

and it is manifest, that the squares of AB, BC, are equal to the square of AC, and twice the square of BC. (1. 47.

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