(I. 31.) CF parallel to AD or BE; then AE is equal to the rectangles AF, CE; and AE is the square of AB; and 1. AF is the rectangle contained by BA, AC, for it is contained by DA, AC, of which AD is equal (Def. 30.) to AB; and 2. CE is contained by AB, BC, for BE is equal to AB; therefore 3. The rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. If, therefore, a straight line, &c. Q.E.D. PROP. 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 aforesaid part. Let the straight line AB be divided into any two parts in the point C; the rectangle AB, BC, is equal to the rectangle 4C, CB, together with the square of BC. C Upon BC describe (I. 46.) the square CDEB, and produce ED to F; draw (I. 31.) AF parallel to CD or BE; then (Constr.) and through 1. and The rectangle AE is equal to the rectangles AD, CE, 2. AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal (Def. 30.) to BC; and 3. AD is contained by AC, CB, for CD is equal to CB; and CE is the square of BC; therefore 4. The rectangle AB, BC, is equal to the rectangle AC, CB, together with the square of BC. If, therefore, a straight line, &c. Q.E.D. PROP. IV.-THEOREM. 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, together with twice the rectangle contained by AC, CB. Upon AB describe (I. 46.) the square ADEB, and join BD; and through C draw (I. 31.) CGF parallel to AD or BE; and through G draw HK parallel to AB or DE. And because CF is parallel to AĎ, and BD falls upon them, (I. 29.) 1. The exterior angle BGC is equal to the interior and opposite angle ADB; but (I. 5.) 2. ADB is equal to the angle ABD, because BA is equal to AD, being sides of a square; wherefore 3. The angle CGB is equal to the angle GBC; and therefore (I. 6.) 4. The side BC is equal to the side CG: but CB is equal (I. 34.) also to GK, and CG to BK; wherefore 5. The figure CGKB is equilateral. It is likewise rectangular; for, since CG is parallel to BK, and CB meets them, therefore (I. 29.) 6. The angles KBC, GCB, are equal to two right angles; and KBC is (Def. 30.) a right angle; wherefore and therefore also (I. 34.) the opposite angles CGK, GKB, are right angles, and 8. The figure CGKB is rectangular. But it is also equilateral, as was demonstrated; wherefore and it is upon the side CB. For the same reason, also, and it is upon the side HG, which is equal (I. 34.) to AC; therefore And because the complement AG is equal (I. 43.) to the complement GE, and that 12. AG is the rectangle contained by AC, CB, for GC is equal (Def. 30.) to CB; therefore also 13. GE is equal to the rectangle AC, CB; wherefore 14. AG, GE, are equal to twice the rectangle AC, CB: and HF, CK, are the squares of AC, CB; wherefore 15. 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 16. 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. COR. From the demonstration, it is manifest that the parallelograms about the diameter of a square are likewise squares. PROP. V.THEOREM. 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 (I. 46.) the square CEFB, join BE, and through D draw (I. 31.) DHG parallel to CE or BF; and through H draw KLM parallel to CB or EF; and also through 4 draw AK parallel to CL or BM. And because the complement CH is equal (I. 43.) to the complement HF, to each of these add DM; therefore 1. The whole CM is equal to the whole DF; but (I. 36.) CM is equal to AL, because AC is equal (Hyp.) to CB; therefore also 2. AL is equal to DF. To each of these add CH, and 3. The whole AH is equal to DF and CH. But AH is the rectangle contained by AD, BD, for DH is equal (II. 4. Cor.) to DB; and DF, together with CH, is the gnomon CMG; therefore 4. The gnomon CMG is equal to the rectangle AD, DB: to each of these add LG, which is equal (II. 4. Cor.) to the square of CD; therefore 5. 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 6. 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. COR. From this proposition it is manifest, that the difference of the squares of two unequal lines, AC, CD, is equal to the rectangle contained by their sum and difference. 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. A C B D Upon CD describe (I. 46.) the square CEFD, join DE, and through B draw (I. 31.) 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 (Hyp.) to CB, (I. 36.) 1. The rectangle AL is equal to CH; but (I. 43.) CH is equal to HF; therefore also 2. AL is equal to HF: to each of these add CM; therefore 3. The whole AM is equal to the gnomon CMG: but AM is the rectangle contained by AD, DB, for DM is equal (II. 4. Cor.) to DB; therefore 4. 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 5. 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 6. 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. VII.-THEOREM. 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. Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC, are equal to twice the rectangle AB, BC, together with the square of AC. Upon AB describe (I. 46.) the square ADEB, and join BD, and through C draw (I.31.) CGF parallel to AD or BE, and through G draw HGK parallel to AB or DE. And because AG is equal (I. 43.) to GE, add to each of them CK; therefore 1. and therefore 2. The whole AK is equal to the whole CE; AK, CE, are double of AK; but AK, CE, are the gnomon AKF, together with the square CK; therefore but 3. The gnomon AKF, together with the square CK, is double of AK; 4. Twice the rectangle AB, BC, is double of AK, for BK is equal (II. 4. Cor.) to BC; therefore 5. The gnomon AKF, together with the square CK, is equal to twice the rectangle AB, BC; to each of these equals add HF, which is equal to the square of AC; therefore 6. The gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC, and the square of AC; |