8. PROPOSITION XXXV.-THEOREM. Parallelograms upon the same base and between the same parallels are equal. Let ABCD, EBCF be the two ms and between the same |3. m Then shall" ABCD = 0 EBCF. upon the same base BC, First, let AD, EF be not terminated in the same pt. then, and B AB, DC are |,.. 4 FDC= LA; · EB, CF are ||3, .. ▲ F = 2 AEB; .. remaining DCF remaining ABE. Also DC AB, and CF = BE; .. ADCF ▲ ABE. (I. 32) (I. 34) (I. 4) Take ▲DCF from trapezium ABCF; and then replace it: afterwards take AABE from the same. The two remainders will be equal. ..□ ABCD = □" EBCF. (Ax. 3) Secondly, let AD, EF be terminated in one pt., E coinciding with D; so that the ' ms become ABCD, DBCF. W B It is plain that these are equal; for each is double of ADBC. Therefore, parallelograms, &c. (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 Produce AG; and make BD, GE each = AC. sq. ACE (I. 46 Now it is easy to shew that the four As DHK, KCE, EG and FBD each ABAC in all respects, .. they = (I. = each other in all respects, and have the sides subtending their right angles all equal; ..the figure DFEK is equilateral. Now, by the same proof, 4 CEK = ▲ GFE, .. ▲ 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 ; But Note. .. sq. DE = sum of sqq. BG, CH. ᎠᏦ = BC; .. sq. on BC = sum of sqq. on BA, AC. The foregoing shews how we may dissect the two small squares so as to fill up with the pieces the area of the larger one. Fo 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. |