8. PROPOSITION XXXV.-Theorem. Parallelograms upon the same base and between the same parallels are equal. Let ABCD, EBCF be the two ☐ms upon the same base BC, and between the same ||. Then shall" ABCD = □" EBCF. First, let AD, EF be not terminated in the same pt. D E ED then, and B B ; · AB, DC are ||3, .. 4 FDC = ▲ A = Also DC :: ΔDCF = Δ ΑΒΕ. F m B A It is plain that these are equal; for each is double of ADBC. Therefore, parallelograms, &c. F (I. 4) Take ADCF from trapezium ABCF; and then replace it: afterwards take ▲ABE from the same. The two remainders will be equal. (Ax. 3) ..□ ABCD = "EBCF. Secondly, let AD, EF be terminated in one pt., E coinciding with D; so that the ms become ABCD, DBCF. (I. 32) (I. 34) (I. 34) Q. E. D. 9. PROPOSITION XLVII.-THEOREM. In a right-angled triangle the square upon the hypotenus equal to the sum of the squares upon the other sides. Let the AABC be rt.- 2a, having the A rt., then the sq. on BC = sum of sqq. on BA, AC. On AB describe the sq. ABFG; and on AC the sq. ACE (I. 46 Produce AG; and make BD, GE each = AC. B But = D H F E Now it is easy to shew that the four ▲s DHK, KCE, EG and FBD each ABAC in all respects, each other in all respects, (I. .. they and have the sides subtending their right angles all equal; .. the figure DFEK is equilateral. Now, by the same proof, ▲ CEK = ▲ GFE, G K .. ▲ FEK = art. 4. Similarly the other 4s of DFEK are rt. .. DFEK is a sq. Now we may superpose DHK on ECK, and DBF on FGE and the remaining part of the two sqq. BG, CH coincides wit the remainder of the sq. DE ; .. sq. DE = sum of sqq. BG, CH. .. sq. on BC= sum of sqq. on BA, AC. Fo Note. The foregoing shews how we may dissect the two smalle squares so as to fill up with the pieces the area of the larger one. this reason chiefly it is inserted; as, fully stated, it is probably at least a long as the usual proof. London: C.F. Hodgson & Son, Printers, Gough Square, Fleet Street, E.C. |