PROPOSITION XLII. THEOREM. 138. The diagonals of a parallelogram bisect each other. A Let the figure A B C E be a parallelogram, and let the diagonals A C and BE cut each other at 0. § 134 § 68 (being alt.-int. £), ZOEA = LOBC; § 68 .:. A A O E=A BOC, § 107 (having a side and two adj. of the one equal respectively to a side and two adj. És of the other). BO= 0 E. and Q. E. D. PROPOSITION XLIII. THEOREM. 139. The diagonals of a rhombus bisect each other at right angles. Let the figure A B C E be a rhombus, having the diagonals AC and BE bisecting each other at 0. $ 128 We are to prove. Z AO E and L A OB rt. £. A E = A B, O E= 0 B, A (= A 0, $ 138 Iden. ..A AO E= A A OB, $ 108 (having three sides of the one equal respectively to three sides of the other); ..ZA O E= L A OB, (being homologous é of equal A); $ 25 (When one straight line meets another straight line so as to make the adj. És equal, each Lis a rt. 2). Q. E. D. PROPOSITION XLIV. THEOREM. 140. Two parallelograms, haring two sides and the included angle of the one equal respectirely to two sides and the included angle of the other, are equal in all respects. B B In the parallelograms A B C D and A' B' C'D', let AB= A' B', A D= A'D', and LA = A'. Apply D A B C D to A' B'C' D', so that A D will fall on and coincide with A' D'. Then A B will fall on A' B', (for ZA= L A', by hyp.), (for A B = A' B', by hyp.). Now, BC and B'C' are both || to A' D' and are drawn through point B'; is the lines B C and B'Ccoincide, § 66 and C falls on B'C' or B'C' produced. In like manner DC and D' Care Il to A' B' and are drawn through the point D'. . DC and D' C' coincide; § 66 ... the point C falls on D' C', or D' C' produced ; .:. C falls on both B' C and D' C"; Q. E, D. 141. COROLLARY. Two rectangles having the same base and altitude are equal; for they may be applied to each other and will coincide. PROPOSITION XLV. THEOREM. 142. The straight line which connects the middle points of the non-parallel sides of a trapezoid is parallel to the parallel sides, and is equal to half their sum. Let S O be the straight line joining the middle points of the non-parallel sides of the trapezoid A BC E. .:. A FOE=AC OH, § 107 (having a side and two adj. € of the one equal respectively to a side and two .. FE=CH, adj. s of the other). and OF= 0 H, Now $ 135 FH= AB, S O is also || to BC, Substitute for A F and BH their equals, A E – FE and BC+CH, and add, observing that CH= FE; then 2 80= A E + BC. ..SO= }(A E + BC). Q. E. D. |