A

с DB

L

H M

Let AB be bisected in C, and produced to D; then the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Upon CD describe the square CEFD, join DE, through B draw K BHG parallel to CE or DF, through H draw KLM parallel to AD or EF, and through A draw AK par

E

allel to CL or DM. AC is equal to CB, therefore the rectangle AL is equal to the rectangle CH, but CH is equal to HF; therefore AL is equal to HF; to each add CM; therefore the whole AM is equal to the gnomon CMG; but AM is the rectangle contained by AD, DB, for DM is equal to DB: therefore the gnomon CMG is equal to the rectangle AD, DB: add to each 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.

VII.-If a straight line be divided into any two parts,

the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part.

H

A

C B

G

K

Let AB be divided into any two parts in C. Then the squares of AB, BC, are equal to twice the rectangle AB, BC, together with the square of AC. Upon AB describe the square ADEB, join BD; through C draw CF parallel to AD or BE cutting BD in G, and through G draw HGK parallel to AB or DE. AG is equal to GE, add to each CK; therefore AK is equal to CE, and AK, CE, are double of AK, but AK, CE are the gnomon AKF and the square CK; therefore the gnomon AKF and the square CK is double cf AK: but twice the rectangle AB, BC, is

D

FE

double of AK, for BK is equal to BC; therefore the gnomon AKF and the square CK are equal to twice the rectangle AB, BC; to each add HF, which is equal to the square of AC, therefore the gnomon AKF, and the squares CK, HF, are equal to twice the rectangle AB, BC, and the square of AC; but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC; therefore the squares of AB and BC are equal to twice the rectangle AB, BC, together with the square of AC.

VIII.—If a straight line be divided into any two parts four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line, which is made up of the whole and that part. This proposition is generally omitted by the student. IX.-If a straight line be divided into two equal, and also into two unequal parts, 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.

Let AB be divided into two equal parts in C, and into two unequal parts in D. Then the squares of AD, DB together, shall be double of the squares of AC, CD.

F.

T

D B

From C draw CE at right angles to AB; make CE equal to AC or CB, and join EA, EB; through D draw DF parallel to CE, meeting EB in F. through F draw FG parallel to BA, and join AF. Because AC is equal to CE, the angle EAC is equal to the angle AEC; and because ACE is a right angle, the two angles AEC, EAC of the triangle are together equal to a right angle; but they are equal; therefore each is half a right angle. Similarly each of CEB, EBC, is half a right angle; and therefore AEB is a right angle; because GEF is half a right angle, and

of

EGF a right angle, for it is equal to the interior and opposite angle ECB; therefore the remaining angle EFG is half a right angle; and GEF is equal to EFG, and the side EG equal to the side GF. 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, therefore the remaining angle BFD is half a right angle; and the angle at B is equal to BFD, and DF to DC. Because AC is equal to CE, the square AC is equal to the square of CE; and the squares of AC, CE double of the square of AC; but the square AE is equal to the squares of AC, CE, since ACE is a right angle; therefore the square of AE is double the square of AC. 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; therefore the square of EF is double of the square of GF; and GF is equal to CD; therefore the square of EF is double of the square of CD, but the square of AE is twice the square of AC; therefore the squares of AE, EF are twice the squares of AC, CD; but the square of AF is equal to the squares of AE, EF, since AEF is a right angle; 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; 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. X.-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 twice the square of half the line bisected, and of the square of the line made up of the half and the part produced.

Let AB be bisected in C, and produced to D; then the squares of AD, DB, are twice the squares of AC,

CD

From C draw CE at right angles to AB, make CE equal to AC or CB, and join AE, EB; through E draw EF parallel to AB, and through DA draw DF parallel to CE. Because

F

B D

G

EF meets the parallels CE, FD, the angles CEF, EFD are equal to two right angles; and the angles BEF, EFD are less than two right angles. But EB, FD, will meet, if produced towards B, D; let them meet in G, and join AG, because AC is equal to CE, therefore the angle CEA is equal to EAC; and ACE is a right angle; therefore each of CEA, EAC is half a right angle. Similarly each of CEB, EBC is half a right angle; therefore AEB is a right angle. And because EBC is half a right angle, DBG is also half a right angle; but BDG is a right angle, since it is equal to the alternate angle DCE; therefore the remaining angle DGB is half a right angle, and is equal to the angle DBG; and BD to DG; because EGF is half a right angle, and the angle at F is a right angle, the remaining angle FEG is half a right angle, and equal to EGF; and GF to FE; because EC is equal to CA; the square of EC is equal to the square of CA; and 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; therefore the square of EA is twice the square of AC. Because GF is equal to EF, the square of GF is equal to the square of EF; 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; therefore the square of EG is twice the square of EF; and EF is equal to CD; therefore the square of EG is double of the square of CD; but the square of EA is double of the square of AC: therefore the squares of EA, EG are twice the squares of AC, CD; but the square of AG is equal to the squares of EA, EG; therefore the square of AG is twice 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 twice the squares of AC, CD;

but DG is equal to DB; therefore the squares of AD, DB are twice the squares of AC, CD.

XI.-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.

E

F

G

H

B

K D

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; bisect AC in E, join BE, produce CA to F, and make EF equal to EB, and upon AF describe the square FGHA. Then A AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. Produce GH to meet CD in K; because the straight line AC is bisected in E, C and produced to F, the rectangle CF, FA, together with the square of AE, is equal to the square of EF; 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, since 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, AE; subtract 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, since FA is equal to FG; and AD is the square of AB; therefore the figure FK is equal to AD; subtract the common part AK, and the remainder FH is equal to the remainder HD; but HD is the rectangle contained by AB, BH, since AB is equal to BD; and FH is the square of AH; therefore the rectangle AB, BH is equal to the square of AH. XII.—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 sub