PROPOSITION XXXII bis. A H F THEOREM.--If in one of two parallel straight lines be taken two points, and through these be drawn two other straight lines figure, Let AB and CD be two Because through the point E is drawn a straight line EH other *I.29.Cor.2. than ED, EH and AB (being prolonged) will* meet. And be cause FI, EH are parallel and DC meets them, the exterior angle 11.29.Cor.1. DFI will+ be equal to the interior and opposite FEH; wherefore FI will be a straight line other than FD, and FI and AB (being $1.29.Cor.2. prolonged) willt meet, and there will be formed a quadrilateral figure; and its opposite sides are* parallel. And by parity of reasoning, the like may be proved in every other instance. Wherefore, universally, if in one of two parallel straight lines &c. Which was to be demonstrated. NOMENCLATURE.—A quadrilateral rectilinear figure, formed (as by the Proposition above) by four lines of which two and two are parallel, is called a parallelogram. Нур. : PROPOSITION XXXIII. See Note. THEOREM.—The opposite sides and angles of every parallelogram are equal to one another; and a diagonal bisects it that is, Let ACDB be a parallelogram, of which BC B Because AB is parallel to CD, and BC meets them, the alterI. 29. nate angles ABC, BCD are* equal to one another. And because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to one another. Wherefore the two tri. angles ABC, DCB have two angles ABC, ACB in the one, equal to two angles BCD, CBD in the other respectively, and one side BC common to the two triangles, and lying between the equal + I. 26. angles in each; therefore their other sides aret equal respectively, viz. the side AB equal to the side CD, and AC to BD; and the angle BAC is equal to the angle BDC; and the triangle ABC is equal to the triangle DCB, and the parallelogram is bisected by BC. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole INTERC. 1. angle ABD is equal to the whole ACD. And in the same way, Cor. 5. if the angular points A and D were joined, might be shown that the opposite sides and angles of the parallelogram were equal, and that the diagonal from A to D likewise bisects the parallelogram. And by parity of reasoning, the like may be proved in every other parallelogram. Wherefore, universally, the opposite sides and angles of every parallelogram &c. Which was to be demonstrated. A G H B CoR. 1. By the help of this Proposition, may be shown how to divide a given straight line AB into any assigned number of See Note. equal parts. For, from one of its extremities A, draw a straight line AC of unlimited length, making with AB any angle less than the sum of two right angles. From the point A in the straight line AC cut off any part as AD; *). 3. and in the same straight line cut off* DE, EF successively equal to AD, till the number of equal parts taken in AC is the same as the number into which it is required to divide AB. Join the extremity of the last (as for instance F) with the extremity B +1 22.Cor. 2. of AB; and at D, E, maket the angles ADG, AEH, respectively 11.27.Cor.2, equal to the angle AFB; wherefore DG, EH will be parallel to FB and to one another, and meet AB as in G, H. Because • I. 12. ADG and GDE are* together equal to two right angles, and AGD and DGH are together equal to two right angles, the * *Interc. 1. four are together* equal to four right angles. But ADG and Cor.5. AGD are togetherf less than two right angles; wherefore GDE + I. 17. and DGH are together greater than two right angles. At G 1.22.Cor.2. then make$ an angle DGI equal to GDA; and because DGI (or GDA) and GDE are together equal to two right angles, and DGH and GDE are together greater than two right angles, *INTERC. 1. DGH is* greater than DGI; and GI, which ist parallel to DE +1.27.Cor.1, and also lies between GH and GD, being prolonged willf meet EH 11.13 Cor.5. between E and H, as in I. Because DGIE is* a parallelogram, • Constr. + I. 33. GI ist equal to DE, and consequentlyť to AD. Also because GI Interc.l. and DE are parallel, and BA meets them; the angles IGHand DAG *1.29.Cor.1. are* equal. And because EH and DG are parallel, and AH meets them ; the angles DGA and IHG are equal. Wherefore the triangles IGH and DAG have the angles IGH and IHG equal to the angles DAG and DGA respectively, and also the side IG + 1. 8. equal to the side DA; consequently the triangles aret equal, and the side GH equal to the side AG. And if GI be prolonged till it meets FB, and from H a straight line be drawn making with HE an angle equal to HEA; in the same way may be shown that HB is equal to GH; and so of any others. . CoR. 2. All the perpendiculars drawn from one of two parallel straight lines towards the other, being prolonged will meet it and be perpendicular to it, and be equal to one another. For because any two such perpendiculars make the two interior angles on the same side of the line (at the points where they meet the straight line from which they are drawn) right angles, they 1.27.Cor.2. are parallel, and the whole of the lines being prolonged will* *1.32 bis. form a parallelogram; wherefore (by Prop. XXXIII above) its opposite sides, among which are the two perpendiculars, are equal, and its opposite angles are also equal. And because each of the angles which the perpendiculars make with the straight line towards which they are drawn is opposite to a right angle, it is a right angle. PROPOSITION XXXIV. THEOREM.--The straight lines which join the extremilies of two equal and parallel straight lines, towards the same parts, are B N joined towards the same parts (that is to say, Join BC. Because AB is parallel to CD, and BC meets them, the alternate • 1. 29, angles ABC, BCD are* equal. And because AB ist equal to + Hyp. CD, and BC common to the two triangles ABC, DCB, the two sides BA, BC are equal to the two CD, CB respectively; and the angle ABC has been shown to be equal to DCB; wherefore the | 1. 4. third side AC isf equal to the third side DB, and the angle ACB to the angle DBC. And because the straight line BC meets the two straight lines AC, BD, and the alternate angles ACB, DBC *J.27.Cor.l. are equal to one another; AC is* parallel to BD. And it has been shown to be also equal to it. And by parity of reasoning, the like may be proved in every other instance. Wherefore, universally, the straight lines which join the extremities of two equal and parallel straight lines, &c. Which was to be demonstrated. PROPOSITION XXXIV. bis. See Note. THEOREM.--Every quadrilateral rectilinear figure, of which the opposite sides are equal to one another, is a parallelogram. Let ACDB be a quadrilateral rectilinear B figure, of which the opposite sides AB, CD are equal to one another, and also the opposite sides AC, BD are equal to one another. ACDB D shall be a parallelogram. Because in the triangles ABC and DCB, AB is equal to CD, and AC to DB, and BC is common, the sides AC and CB of the one are equal to the sides DB and BC of the other respectively, and also the third side AB of the one is equal to the third side DC 1.8. of the other; wherefore the angle ACB of the one ist equal to the angle DBC of the other, and the angle ABC to the angle DCB. But because the angles ACB, DBC are equal to one 11.27.Cor.l. another, and they are alternate angles, AC ist parallel to BD; and because the angles ABC, DCB are equal to one another, and they are alternate angles, AB is parallel to CD; wherefore 'I. 32 bis. ACDB is* a parallelogram. Nom. And by parity of reasoning the like may be proved of every other quadrilateral rectilinear figure, of which the 'opposite sides are equal to one another. Wherefore, universally, a quadrilateral rectilinear figure, of which &c. Which was to be demonstrated. CoR. 1. In any quadrilateral rectilinear figure of which the opposite sides are equal, if one of the angles be a right angle, all the angles shall be right angles. For, because the quadrilateral figure (by the Prop. above) is a * I. 33. parallelogram, the opposite angles are equal to one another; wherefore if one of the angles be a right angle, the angle opposite to it will also be a right angle, and the two together will be equal to two right angles. And because all the angles of every quadri+1.32.Cor.1. lateral rectilinear figure aret together equal to four right angles, the two remaining angles will be together equal to two right 11. 33. angles. And because they are opposite angles, they are equal to one another; therefore each of them will be a right angle. Cor. 2. If one of the angles be not a right angle, none of the angles shall be right angles; but there shall be two angles greater than right angles, and two less. For if one of the angles be greater than a right angle, the angle opposite to it will also be greater; and the two will be together greater than two right angles; and the two remaining angles will be together less than two right angles; and because they are equal to one another, each of them will be less than a right angle. And in the same way, if one angle and its opposite be less than right angles, the others will be greater. *Cor. 1 NOMENCLATURE 1. A parallelogram of which the angles are* above. right angles, is called a rectangle. A rectangle is said to be made by any two of the straight lines between which is one of its right angles. Or for brevity, the rectangle made by two straight lines AB, BC, is called the rectangle AB, BC ; or more briefly still, the rectangle ABC. Nom. 2. A rectangle of which all the sides are equal, is called a square. Nom. 3. A rectangle of which only the opposite sides are equal, is called an oblong. Nom. 4. A parallelogram of which all the sides Cor. 2 are equal, but the angles aret not right angles, is above. called a rhombus. |