Otherwise thus: The square of AB is equal to double the rectangle Fig. 16. ACB, together with the squares of AC and CB; add to both the square of CB, and the square of AB together with the square of CB is equal to double the rectangle ACB together with the square of AC and double the square of CB; but double the rectangle ACB with double the square of CB is equal to double the rectangle ABC, therefore the squares of AB and CB are together equal to double the rectangle ABC together with the square of AC. Cor. Hence it is evident that the excess of the sum Fig. 16. of the squares of two lines AB and CB above the double rectangle under them ABC is equal to the square of the difference between them AC. PROP. VIII. THEOR. If a right line (AC) be divided into any two parts Fig. 17. (in B), the square of the sum of the whole line (AC) and either segment (BC) is equal to four times the rectangle under the whole line (AC) and that segment (BC) together with the square of the other segment (AB). 1. Produce AC till CD is equal to BC (1), on AD de- (1) Post, 1. scribe the square ARZD (2), and through the points & P. 3. B. B and C draw BS and CV parallel to AR, having (2) Prop. drawn RD draw through the points G and K, EH 46. B. 1. and LP parallel to AD. Because SV is equal to BC (3), BC to CD (4), and (3) Prop.34 CD to VZ, SV and VZ are equal, and therefore the B. 1. (4) Constr. rectangles SG and VH are equal (5), but VH and AG (5) Prop.36 are also equal (6), therefore SG is equal to AG; and B. 1. because FG is equal to BC(3), FG and CD are equal, (6) Prop.43 and therefore the square FO is equal to the square CH; but also EK and KV are equal (6), to these equals if the equals CH and FO be added, EK and CH together shall be equal to SG, and therefore to AG; therefore B. 1. (7) Cor. prop. 43. AG, SG, and VH, together with EK and CH are four times AG, but AG, SG, and VH, together with EK, CH, and the square LS are equal to the square AZ, therefore AG four times taken together with LS is equal to AZ. But AG is the rectangle under AC and BC, because CG is equal to CD (7), and therefore to BC (8), and LS is the square of AB, because AB and RS are (8) Constr. (9) Prop.34 equal (9). B. 1. B. 1. Fig. 18. Otherwise thus: Produce AC till CD is equal to BC, the square of AD is equal to the square of AC and CD, together with double the rectangle ACD, that is, because BC and (1) Constr. CD are equal (1), to the squares of AC and BC, together with double the rectangle ACB, but the squares of AC and BC are equal to double the rectangle ACB (2) Frop. 7. together with the square of AB (2), therefore the square of AD is equal to the rectangle ACB four times taken together with the square of AB. B. 2. Fig. 19. PROP. IX. THEOR. If a right line be cut into equal parts (in C) and into unequal (in D), the sum af the squares of the unequal parts (AD and DB) is equal to double the sum of the squares of the half (AC) and of the intermediate part (CD). From the point C draw CE perpendicular to AB (1)Prop.11 and equal to either AC or CB (1), join AE and EB, 3. B. 1. and through D draw DF parallel to CE, and through F draw FG parallel to CD and join FA. Because the angle ACE is a right angle,and the sides (2) Constr. AC and CE are equal (2), CEA is half a right angle (3) Cor. (3), in the same manner it can be demonstrated that 5. Prop.52. CEB is half a right angle, therefore AEB is a right angle; on account of the parallels GF and CD, the (4) Prop.29 angle EGF is equal ECB (4), therefore EGF is a right angle, but GEF is half a right angle, therefore GFE is also half a right angle, and therefore GE and GF are B. 1. B. 1. B. 1. equal (5); likewise FDB is a right angle, because it is (5) Prop. 6. Otherwise thus: B. 1. The square of AD is equal to the squares of AC Fig. 20. and CD, together with double the rectangle ACD (1), (1)Prop.4. or, because AC and CB are equal, to double the rect- B. 2. angle BCD; add to both the square of BD, and the squares of AD and DB are equal to the squares of AC, CD, DB together with double the rectangle BCD, but double the rectangle BCD, with the square of DB, is equal to the squares of CB and CD (2), or, because (2) Prop. 7. AC and CB are equal, to the squares of AC and CD, B. 2. therefore the sum of the squares of AD and DB is equal to double the sum of the squares of CD and AC. This proposition, is not be bandb. Fig. 21. & 3.B. 1. (2) Prop.30 B. 1. PROP. X. THEOR. If a right line (AB) be bisected (in C) and produced to any point (D), the square of the whole line thus produced (AD), together with the square of the produced part (BD) is equal to double the square of the line (CD) made up of the half and produced part together with double the square of half the given line. From the point C draw CE perpendicular to AB (1)Prop. 11 and equal to either CA or CB (1), join AE, and draw through the point E, EF parallel to AB (2), and through D, DF parallel to CE; and because the angles CEF and DFE are equal to two right angles (3) Prop.29 on account of the parallel lines CE and DF (3), the angles BEF and DFE are less than two right angles, therefore the lines EB and FD, if produced, shall (4) 4x. 12. meet (4), let them meet in G and draw GA. (5) Constr. B. 1. P. 32. B. 1. Because CA and CE are equal (5), and the angle C a right angle (5), the angle CEA is half a right angle (6) Cor. 3. (6), in the same manner it is proved that CEB is half a right angle, therefore AEB is a right angle, and because DG and EC are parallel (5), the alternate angles GDB, ECB are equal, therefore GDB is a right angle; (7) Prop.15 also the angles DBG and EBC are equal (7), but EBC is half a right angle, therefore DBG is half a right angle and also DGB, and therefore the sides DB and (8) Prop.6. DG are equal (8), and because EGF is half a right angle, and the angle at F right, being equal to its op(9) Prop.34 posite C (9), FEG is half a right angle, and therefore the sides EF and FG are equal. B. 1. B. 1. B. 1. Because AC and CE are equal, and the angle ACE right, the square of AE is double the square of AC; and because GF and FE are equal, and the angle F right, the square of GE is double the square of EF, but EF and CD are equal (9), therefore the square of GE is double the square of CD; the square of AE is also double the square of AC, therefore the squares of AE and EG are together double the squares of AC and CD: but the square of AG is equal to the squares of AE and EG, and is therefore double the squares of AC and CD, and the squares of AD and DG are equal to the square of AG, and therefore double the squares of AC and CD, but BD and DG are equal, and therefore the squares of AD and DB are double the squares of AC and CD. Otherwise thus: The square of AD is equal to the squares of AC Fig. 22. and CD, together with double the rectangle ACD (1), (1) Prop.4. or double the rectangle DCB, because AC and CB are B. 2. equal. But double the rectangle DCB, together with the square of DB is equal to the squares of DC, and CB or AC (2). Therefore the square of AD, together (2) Prop.7. with the square of DB, is equal to double the square of AC and double the square of CD. Cor. The sum of the squares of two lines is equal to double the square of half their sum and double the square of half their difference. For CD is half the sum of the lines AD and BD, and AC is half their difference. B. 2. PROP. XI. PROB. To divide a given finite right line (AB), so that the Fig. 23. rectangle under the whole line and one segment shall be equal to the square of the other segment. From the point A erect AC perpendicular and equal to the given line AB, bisect it in E, join EB, produce CA until EF is equal to EB, and in the given line AB take AH equal to AF; the square of AH is equal to the rectangle under the other segment HB and the whole line AB. Complete the square of AB, draw through H the right line GK parallel to AC, and through F the right line FG parallel to AB. Because CA is bisected in E and AF is added to it, the rectangle under CF and FA, together with the |