but MP is equal to PL, because they are the complements of the parallelogram ML; (1. 43)

and AG is equal to RF;

therefore the four rectangles AG, MP, PL, RF, are equal to one another, and so are quadruple of one of them, AG.

And it was demonstrated that the four CK, BN, GR, and RN, are quadruple of CK;

therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK.

And because AK is the rectangle contained by AB, BC, for BK is equal to BC,

therefore four times the rectangle AB, BC, is quadruple of AK ; but the gnomon AOH was demonstrated to be quadruple of AK;

therefore four times the rectangle AB, BC, is equal to the gnomon АОН;

to each of these add XH, which is equal to the square of AC; (11.4, Cor.) therefore four times the rectangle AB, BC, together with the square of AC, is equal to the gnomon AOH, and the square XH;

but the gnomon AOH and XH make up the whole figure AEFD, which is the square of AD;

therefore four times the rectangle AB, BC, together with the square of AC, is equal to the square of AD,

that is, of AB and BC added together in one straight

line.

Wherefore, if a straight line, &c.

Q. E. D.

PROP. IX.—THEOREM.

If a straight line be divided into two equal and also into two unequal parts;

then the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.

(Reference-Prop. r. 3, 5, 6, 11, 29, 31, 32, 34, 47.)

Let the straight line AB be divided into two equal parts in the point C, and into unequal parts in the point D.

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

From the point C draw CE at right angles to AB. (1. 11)
and make it equal to AC or CB, (1. 3) and join EA, EB;
through D draw DF parallel to CE; (1. 31) meeting GB in F;
and through F draw FG parallel to AB; and join AF.

DEMONSTRATION

Then, because AC is equal to CE, (constr.)

the angle EAC is equal to the angle AEC; (1. 5)

and because the angle ACE is a right angle,

the angles AEC, EAC, together make one right angle; (1. 32) and they are equal to one another;

therefore each of the angles AEC, EAC, is half a right angle.

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 the angle GEF is half a right angle,

and EGF a right angle, for it is equal to the interior and opposite angle ECB, (1. 29)

the remaining angle EFG is half a right angle;

therefore the angle GEF is equal to the angle EFG,

wherefore the side EG is 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)

the remaining angle BFD is half a right angle;

therefore the angle at B is equal to the angle BFD, wherefore the side DF is 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 squares of AC; but because ACE is a right angle

the square of EA is equal to the squares of AC, CE, (1. 47) therefore the square of EA 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 EA is likewise double of the square of AC; therefore the squares of EA, EF, are double of the squares of AC, CD. And because AEF is a right angle,

the square of AF is equal to the squares of AE, EF, (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 squares of AF, because 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.

Therefore if a straight line, &c.

Q. E. D.

PROP. X.-THEOREM.

If a straight line be bisected, and produced to any point;

then 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.

(References-Prop. 1. 5, 6, 11, 15, 29, 31, 32, 34, 46 Cor. 47.)

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 it 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 EC, FD, the angles CEF, EFD, are equal to two right angles; (1. 29) 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, do meet if produced far enough;' (ax. 12)

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

let them meet in G, and join AG.

DEMONSTRATION

Then, because AC is equal to CE,

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, the angle DBG which is vertically opposite is also half a right angle, (1. 15)

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 augle, and is there

fore 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 that the angle at F is a right angle, because it is equal to the oppo

site angle ECD, (1. 34)

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 FE.

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 to 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.

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.

(References

Prop. I. 3, 10, 46, 47; II. 6.)

Let AB be the given straight line.

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

E