19. Given the sum of the hypotenuse and a side of a right-angled triangle, and also the remaining side, to construct it.

20. Given the sum of the sides of a triangle and the angles at the base, to construct it.

.

21. If two lines be drawn from the extremities of the base of a triangle to bisect the opposite sides, the line joining their intersection with the vertex, if produced, will bisect the base.

22. If two polygons be constructed on the same side of the same base, the sum of the sides of the interior polygon, if it be concave internally, is less than the sum of the sides of the exterior figure.

SECOND BOOK.

DEFINITIONS.

1. If a straight line be divided or cut at any point, the point is called a point of section.

2. The distance of the point of section from the middle of the line, is called the mean distance.

3. If the point of section lie between the extremities of the line, it is said to be cut internally; and if beyond one of the extremities, it is said to be cut externally.

4. The distances from a point of section of a line to its extremities, are called segments; which are said to be internal or external, according as the line is cut internally or externally.

5. The rectangle under, or contained by, two lines, is a rectangle, of which these lines, or lines equal to them, are two adjacent sides. The rectangle under two lines, as AB and BC, is sometimes concisely expressed thus, AB BC; or if A and B denote the lines, by A B. So the rectangle under a line, which is the sum of two lines A and B, and a third line C, is expressed thus (A + B) C; that under the excess of A above B, and another line C, by (AB) C;

and that under a line equal to A+B, and another equal to A - B, by (A + B) (A — B).

6. When a line is cut into two segments, so that the rectangle under the whole line and one of the segments is equal to the square of the other segment, it is said to be cut medially, or in medial section.

7. Any of the parallelograms about a diagonal of any parallelogram, together with the two complements, is called

a gnomon.

Thus the parallelogram HG, together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon.

H

B

E

If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Let A and BC be two straight lines, and let BC be divided into any parts in the points D, E; the rectangle contained by the straight lines A, BC, is equal

to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC.

D E C

K L H

A

From the point B draw (I. 11) BF at F right angles to BC, and make BG equal to A (I. 3); and through G draw GH parallel to BC (I. 31); and through D, E, C, draw (I. 31) DK, EL, CH, parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, because DK, that is, BG (I. 34), is equal to A; and in like manner the rectangle EH is contained by A, EC; therefore the rectangle contained by A, BC, is equal to the several rectangles contained by A, BD, and by A, DE, and also by A, ĚC.

If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts are together equal to the square of the whole line.

Let the straight line AB be divided into any two parts in the point C; the rectangle contained by AB, BC, together with the rectangle AB AC, shall be equal to

the square of AB.

Upon AB describe (I. 46) the square ADEB, and through C draw CF, parallel to AD or BE (I. 31); then AE is equal to the rectangles AF, D CE; and AE is the square of AB; and AF is the rectangle contained by BA, AC; for it is contained by DA, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE is equal to AB; therefore the rectangle contained by AB, AC, together with the rectangle AB BC, is equal to the square of AB.

PROPOSITION III. THEOREM.

If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the rectangle contained by the two parts, together with the square of the foresaid part.

Let the straight line AB be divided into two parts in the point C; the rectangle AB BC, is equal to the rectangle AC CB, together with the square of BC.

C

B

Upon BC describe (I. 46) the square CDEB, and produce ED to F, and through A draw AF A parallel to CD or BE (I.31); then the rectangle AE is equal to the rectangles AD, CE. Now, AE is the rectangle contained by AB, BC, for it is contained by

D

AB, BE, of which BE is equal to BC; and AD is contained by AC, CB, for CD is equal to CB; and DB is the square of BC; therefore the rectangle AB BC is equal to the rectangle AC CB, together with the square of BC.

If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC CB.

A

H

C B

G

B

F E

Upon AB describe the square ADEB (I. 46), and join BD, and through C draw CGF parallel to AD or BE (I. 31), and through G draw HK parallel to AB or DE. And because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal to the interior and opposite angle ADB (I. 29); but ADB is equal to the angle ABD (I. 5), because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle GBC; and therefore the side BC is equal to the side ČG (I. 6); but CB is equal also to GK (I. 34), and CG to BK; wherefore the figure CGKB is equilateral. It is likewise rectangular; for the angle CBK being a right angle, the other angles of the parallelogram CGKB are also right angles (I. 46, Cor.); wherefore CGKB is a square, and it is upon the side CB. For the same reason, HF also is a square, and it is upon the side HG, which is equal to AC; therefore HF, CK, are the squares of AC, CB. And because the complement AG is equal (I. 43) to the complement GE, and because AG is the rectangle contained by AC, CG, that is, by AC, CB; GE is also equal to the rectangle AC CB; wherefore AG, GE, are equal to twice the rectangle AC CB. And HF, CK, are the squares of AC, CB; wherefore the four figures HF, CK, AG, GE, are equal to the squares of AC, CB, and to twice the rectangle AC CB.~ But HF, CK, AG, GE, make up the whole figure ADEB, which is the square of AB; therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC CB.

COR. From the demonstration, it is manifest that the parallelograms about the diagonal of a square are likewise squares.

If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D; the rectangle AD DB, together with the square of CD, is equal to the square of CB.

Upon CB describe the square CEFB (I. 46); join BE, and through D draw DHG parallel to CE or BF (I. 31); and through H draw KLM parallel to A

CB or EF; and also through A draw
AK parallel to CL or BM.

And be- K

cause the complement CH is equal to the complement HF (I. 43), to each of

D B

H M

these add DM; therefore the whole CM is equal to the whole DF; but CM is equal to AL (I. 36), because AC is equal to CB; therefore also AL is equal to DF. To each of these add CH, and the whole AH is equal to DF and CH; but AH is the rectangle contained by AD, DB, for DH is equal to DB (II. 4, Cor.); and DF, together with CH, is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD DB. To each of these add LG, which is equal to the square of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD DB, together with the square of CD. But the gnomon CMG and LG make up the whole figures CEFB, which is the square of CB; therefore the rectangle AD DB, together with the square of CD, is equal to the square of CB.

COR. From this proposition it is manifest, that the difference of the squares of two unequal lines BC, CD, is equal to the rectangle contained by their sum and difference.

For BC2 AD DB + CD2, and taking CD2 from both these equals, there remains BC2 CD2 = AD-DB= (BC+CD) (BC-CD), for BC+CD=AC+CD=AD.