138. The diagonals of a parallelogram bisect each other. Let the figure A B C E be a parallelogram, and let the diagonals AC and B E cut each other at 0. We are to prove A0 = 0 C, and BO= 0 E. , = ..A AO E= A BOC, $ 107 (having a side and two adj. & of the one equal respectively to a side and two adj. s of the other). ..AO=OC, and BO= 0 E. Q. E. D. 139. The diagonals of a rhombus bisect each other at right angles. А E B Let the figure A B C E be a rhombus, having the diagonals AC and BE bisecting each other at 0. § 108 .:.Δ ΑΟ Ε =Δ Α Ο Β, (having three sides of the one equal respectively to three sides of the other); ..LAOE-LAOB, (being homologous É of equal A); ..LAO E and ZA OB are rt. [. § 25 (When one straight line meets another straight line so as to make the adj. És equal, each Zis a rt. Z). Q. E. D. PROPOSITION XLIV. THEOREM. 140. Two parallelograms, having two sides and the included angle of the one equal respectively to two sides and the included angle of the other, are equal in all respects. В! B с A Α' DI In the parallelograms A B C D and A' B' C'D', let A B = A' B', A D= A'D', and ZA= A'. Apply D A B C D to O 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 LA= 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'; .. the lines B C and B'C' coincide, and C falls on B'C' or B'C' produced. In like manner D C and D' Care || 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"; § 66 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. 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 B C E. (being vertical 6 ). .:. A FOE=AC OH, $ 107 (having a side and two adj. És of the one equal respectively to a side and two adj. es of the other). .. FE=CH, 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. |