opposite sides of AB, make with it at the point A, the adjacent angles equal to two right angles; therefore AC is in the same straight line (2. 1.) with AG: for the same reason, AB and AH are in the same straight line; and because the angle DBC is equal to the angle FBA, each of them being a right angle (Def. 5. 2.), add to each the angle ABC, and the whole angle DBA is equal to the whole FBC; and be cause the two sides AB, BD are equal to the two BF, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal (4. 1.) to the base CF, and the triangle ABD is equivalent to the triangle FBC; now the parallelogram BL is double (15. 2.) of the triangle ABD, because they are upon the same base BD, and between the same parallels BD, AL; and the square BG is double (15. 2.) of the triangle FBC, because these also are upon the same base BF, and between the same parallels BF, CG. But the doubles of equals are equal to one another; therefore the parallelogram BL, is equal to the square BG; and in the same manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square CH; therefore the whole square BDEC is equal to the two squares BG, CH; and the square BDEC is described upon the straight line BC, and the squares BG, CH upon AB, AC; therefore the square upon the side BC, is equal to the squares upon the sides AB, AC. Therefore, in any right angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. Which was to be proved.

PROP. XVIII. THEOR.

If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle.

If the square described upon BC, one of the sides of the triangle

ABC, be equal to the squares upon the other sides AB, AC, the angle BAC is a right angle.

A

From the point A, draw AD at right angles (9. 1.) to AC, and make AD equal to AB, and join CD; then, because AD is equal to AB, the square of AD is equal to the square of AB; 'to each of these add the square of AC; there'fore the squares of AD, AC are equal to the squares of AB, AC; but the square of CD is equal (17. 2.) to the squares of AD, AC, because DAC is a right angle; and the square of BC, by hypothesis, is equal to the squares of AB, AC; therefore the square of DC, is equal to the square of BC: and therefore, also, the side DC is equal to the side BC. And because the side AD is equal to AB, and AC common to the two triangles DAC, BAC, the two sides AD, AC are equal to the two sides AB, AC, and the base DC is equal to the base BC; therefore the angle DAC is equal (7. 1.) to the angle BAC; but DAC is a right angle; therefore also BAC is a right angle. Therefore, If the square, &c.

PROP. XIX. THEOR.

B

The complements of the parallelograms which are about the diagonal of any parallelogram, are equal to one another.

Let ABCD be a parallelogram, of which the diagonal is AC, and EH, FG the parallelograms about AC, that is, through which AC passes, and BH, DK, the other parallelograms which make up the whole figure ABCD, which are therefore called the complements; the complement BK is equal to the complement E

DK.

B

A H

G

D

K

F

Because ABCD is a parallelogram, and AC its diagonal, the triangle ABC is equal (Cor. 9. 2.) to the triangle ADC; and because AEKḤ is a parallelogram, the diagonal of which is AK, the triangle AEK is equal to the triangle AHK; for the same reason, the triangle KGC is equal to the triangle KFC, then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC equal to the triangle KFC; the triangle AEK, together with the triangle KGC, is equal to the triangle AHK, together with the triangle KFC; but the whole triangle ABC, is equal to the whole triangle ADC; therefore the remaining complement BK, is equal to the remaining complement DK. Therefore, the complements of the parallelograms which are about the diagonal of any parallelogram, are equal to one another. Which was to be proved.

PROP. XX. THEOR.

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.

H

A

G

C B

K

团

Upon AB describe (16. 2.) the square ADEB, and join BD, and through C draw (4. 2.) CGF parallel to AD or BE, and through G draw HK parallel to AB or DE; and because CF is parallel to AD, and BD falls upon them, the angle BGC is equal (3. 2.) to the corresponding angle ADB; but ADB is equal (5. 1.) to the angle ABD, 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 (6. 1.) to the side CG; but CB is equal also (9. 2.) to GK, and CG to BK ; wherefore the figure CGKB is equilateral it is likewise rectangular; for CG is parallel to BK, and CB meets them, the angles KBC, GCB are therefore equal to two right angles; and KBC is a right angle, wherefore GCB is a right angle; and therefore also the angles CGK, GKB opposite to these are right angles, and CGKB is rectangular; but it is also equiF E lateral, as was demonstrated; wherefore it 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 (19. 2.) to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore 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. Wherefore, If a straight line, &c. Q. E. D.

D

COR. The parallelograms about the diagonal of a square are likewise squares.

PROP. XXI. THEOR.

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 of half the line.

square

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 (16. 2.) the square CEFB, join BE, and through D draw (4. 2.) DHG parallel to CE or BF; and through H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM; and because the complement CH is equal (19. 2.) to the complement HF, to each of these add DM, therefore the whole CM is equal to the whole DF; but CM is equal (14. 2.) to AL, 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 (Cor. 20. 2.) to DB; and DF together with CH is the gnomon CMG; therefore the gnomon CMG is equal to

K

C

D B

L

HM

E

G F

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 figure 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. Wherefore, If a straight line, &c. Q. E. D.

PROP. XXII. THEOR.

If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D; the rectangle AD, DB, together with the square of CB, is equal to the square of CD.

Upon CD describe (16.2.) the square CEFD, join DE, and through B draw (4. 2.) BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK parallel to CL or DM and because AC is equal to CB, the rectangle AL is equal

(14.2.) to CH; but CH is equal A (19. 2.) to HF; therefore also

C

B D

AL is equal to HF to each of

these add CM; therefore the. whole AM is equal to the gnomon CMG and AM is the rectangle contained by AD, BD, for DM is equal (Cor. 20. 2.) to DB: therefore the gnomon CMG is equal to the rectangle AD, DB; add to each of these LG,

which is equal to the square of CB; therefore the rectangle AD DB, together with the square of CB, is equal to the gnomon CMG and the figure LG: but the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXIII. THEOR.

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; the squares of AD, DB are double of the squares of AC, CD.

From the point C draw (9. 1.) CE at right angles to AB, and make it equal to AC or CB, and join AE, EB; through E draw (4. 2.) EF parallel to AB, and through D draw DF parallel to CE; and because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal (1 Cor. 3. 2.) to two right angles; 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, do meet (13. 2.) if produced far enough; therefore EB, FD shall meet, if produced, toward BD; let them meet in G, and join AG; then because AC is equal to CE, the angle CEA is equal (5. 1.) to the angle EAC; and the angle ACE is