LECTURE XIV. THE PARALLELOGRAM.-PROPOSITIONS 33-34. I. 33. PROPOSITION XXXIII. shows the development of the parallelogram from parallel straight lines. It teaches us that the straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel. Let AB and CD be equal and parallel lines joined by AC and BD, we have to show that AC is equal and parallel to BD. I. To show that AC is equal to BD. In order to show that lines are equal, we must generally prove either that they are radii of the same circle or sides of equal triangles. The figure ABDC can be divided into two triangles by joining any two of the angular points. Let us join BC. Then in the triangles ABC, BCD, we have ABCD by hypothe B sis and CB common. But do we know anything about the included angle? Since AB and CD are parallels and BC meets them, the alternate angles are equal, viz. ▲ ABC = ▲ BCD. Therefore the triangles ABC and BCD are equal in all respects, and AC is equal to BD. II. It remains to show that AC is also parallel to BD. AC and BD are met by the line BC, are the alternate angles equal? Yes; for the alternate angles made by BC with AC and BD are also angles of the triangles, which we have just shown to be equal. Therefore AC is parallel to BD. And the figure ABDC is a parallelogram, that is a foursided figure whose opposite sides are parallel, and it is formed by joining the extremities of equal and parallel straight lines. SUMMARY OF PROPOSITION XXXIII., THEOREM 23. The straight lines which join the extremities of equal and parallel straight lines towards the same parts are themselves equal and parallel. .. A ABC = A BCD (I. 4), and AC = BD. II. To show that AB is || to CD. BC falls on AC and BD and makes Z ACB = ▲ CBD (I. 4) ; .. AC is to BD (I. 29). Euclid in the definitions describes four kinds of quadrilateral figures having their opposite sides equal—the square, the rhombus, the rhomboid, and the oblong, more generally spoken of as the rectangle. It may easily be proved that all of these have their opposite sides parallel as well as equal. Let ABDC be any of these figures, e.g. a rhomboid. Join BC. Then the three sides of the two triangles into which it is divided are equal, and the triangles are equal in all respects (I. 8), and hence their angles are equal. Therefore BC falling on AB, CD, and also on AC, BD, in each case makes the alternate angles equal, and therefore each of these pairs of sides are parallels. Hence all of these figures may be classed together as parallelograms. I. 34. In the next proposition we have to show that the opposite sides and angles of a parallelogram are equal, and that the diameter bisects it. The diameter, or diagonal, of a parallelogram is the line which joins any pair of its opposite angles, dividing it into two triangles. Let ABDC be a parallelogram, we have to show that AB = CD, AC = BD, and that ▲ ABC = ▲ BCD, which must be the case if the diagonal bisects the figure. It will be better to invert the order in proving the three points of this theorem, for if the triangles are equal it will follow that their sides are equal, and consequently the whole theorem will be proved. What do we know about these two triangles, as to their sides and as to their angles? BC is common to the two triangles. Since AB is parallel to CD, and BC meets them, the alternate angles are equal, i.e. ▲ ABC = 4 BCD. A B D Again, since AC is parallel to BD and BC meets them, the alternate angles ACB, CBD are equal. Then in the two triangles we have two angles and the side between them equal, therefore the triangles are equal in all respects (I. 26, pt. 1). Since the triangles are equal in area, the diameter bisects the parallelogram. Since the sides of the triangles are equal, each to each, AC BD, and AB = CD, and the opposite sides of the parallelograms are equal. = Since the angles of the triangles are equal, each to each, < CAB = BDC, and adding the other equal angles, ▲ ACB + 2 BCD = 4 ABC + ≤ CBD, i.e. ▲ ACD = 2 ABD. Therefore the opposite angles of the parallelogram ABDC are equal. NOTE. This is the first proposition in which we have made use of all the inferences which can be drawn from the equality of any two triangles; in former propositions we have shown triangles to be equal, in order to prove the equality of certain pairs of angles, or of sides. SUMMARY OF PROPOSITION XXXIV., THEOREM 24. The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it. follows: B Cons.-Nil. Proof-Alternate angles made by BC with both pairs of parallels are equal. BC is common to the two triangles ABC, BCD. .. ▲ ABC = ▲ BCD (I. 26). It I. AB = CD, AC = BD, .. opposite sides of gram are equal. II. ▲ ABC = BCD in area, .. diameter bisects ||gram ABDC. III. / CAB= / BDC; ▲ ACD (ACB + BCD) = 2 ABD (ABC + CBD) .. opposite angles of gram are equal. Let us briefly recapitulate what we have learnt about the parallelogram. It is formed by joining the ends of equal and parallel straight lines towards the same parts. Its opposite sides are equal and parallel. Its opposite angles are equal to each other, all its interior angles are together equal to four right angles and to its exterior angles if the sides be produced. There are four kinds of parallelograms :— The square, whose sides and angles are all equal. The rhombus, whose sides are all equal, but only the opposite angles. The rhomboid, whose opposite sides and angles are equal. The oblong or rectangle, whose angles are all equal, but opposite sides only. Of these figures two, the square and the rectangle, have all their angles right angles. |