PROPOSITION XXX. THEOREM. Right Lines parallel to one and the fame Right Line, are alfo parallel between themselves. LETAB and CD be Right Lines, each of which is parallel to the Right Line E F. I fay, AB is alfo parallel to CD. For let the Right Line G K fail upon them. Then because the Right Line GK falls upon the 29 of this. parallel Right Lines AB, E F, the Angle A G H is* equal to the Angle G H F; and because the Right Line GK falls upon the parallel Right Lines EF, +29 of this. CD, the Angle GHF is equal to the Angle GKD+. But it has been proved that the Angle A G K is also equal to the Angle GHF. Therefore A G K is equal G K D, and they are alternate Angles; whence 27 of this. A B is parallel to C D‡. And fo, Right Lines, parallel to one and the fame Right Lines, are parallel between themfelves; which was to be demonftrated. PROPOSITION XXXI. PROBLEM. To draw a Right Line through a given Point parallel to a given Right Line. LETA be a Point given, and BC a Right Line given. It is required to draw a Right Line thro' the Point A, parallel to the Right Line BC. Affume any Point D in B C, and join AD; then 23 of this. make an Angle D A E, at the Point A, with the Line D A, equal to the Angle A DC, and produce E A ftrait forwards to F. Then, because the Right Line A D, falling on two Right Lines B C, EF, makes the alternate Angles EAD, ADC, equal between themselves, E F fhall +27 of this. be parallel to BC. Therefore, the Right Line EA F, is drawn from the given Point A, parallel to the given Right Line BC; which was to be done. PRO. PROPOSITION XXXII. THEOREM. If one Side of any Triangle be produced, the outward Angle is equal to both the inward and oppofite Angles: and the three inward Angles of a Triangle are equal to two Right Angles. L ET ABC be a Triangle, one of whofe Sides BC is produced to D. I fay the outward Angle ACD is equal to the two inward and oppofite Angle C AB, A B C; and the three inward Angles of the Triangle, viz. ABC, BCA, CAB, are equal to two Right Angles. For let CE be drawn thro' the Point C, parallel 31 of ibin to the Right Line A B. Then, because A B is parallel to CE, and A C falls upon them, the alternate Angles BAC, ACE, are equal between themselves. † 29 of sbis. is equal to the inward and oppofite one ABC; but || 29 of this Coroll. 1. All the three Angles of any one Triangle taken together, are equal to all the three Angles of any other Triangle, taken together. Caroll. 2. If two Angles of any one Triangle, either feparately, or taken together, be equal to two Angles of any other Triangle; then the remaining Angle of of the one Triangle will be equal to the remaining 'Angle of the other. Coroll. 3. If one Angle of a Triangle be a Right Angle, the other two Angles together make one Right Angle. Coroll. 4. If the Angle included between the equal Legs of an Ifoceles Triangle be a Right one, each of the other Angles at the Bafe will be half a Right Angle. Corall. 5. Any Angle in an Equilateral Triangle is equal to one Third of two Right Angles, or two Thirds of one Right Angle. Coroll. 6. Hence it appears, that if one Angle of any Triangle be equal to the other two, that is a Right one; because that the Angle adjacent to this Right one, is equal to the other two. But when adjacent Angles are equal, they are neceffarily Right ones. THEOREM I. All the inward Angles of any Right-lin❜d Figure whatsoever, make twice as many Right Angles, abating four, as the Figure has Sides. FOR any Right-lin'd Figure may be refolved into as many Triangles, abating two, as it hath Sides. For Example, if a Figure has four Sides, it may be refolved into two Triangles: If a Figure hath five Sides, it may be refolved into three Triangles; if fix, into four; and so on. Wherefore (by Prop. XXXII.) the fo Angles of all thefe Triangles are equal to twice as many Right Angles, as there are Triangles: But the Angles of all the Triangles are equal to the inward Angles of the Figure. Therefore all the inward Angles of the Figure are equal to twice as many Right Angles as there are Triangles, that is, twice as many Right Angles, taking away four, as the Figure hath Sides. ᎳᎳ Ꭰ. THE THEOREM II. All the outward Angles of any Right-lin❜d Figure, together make four Right Angles. FOR the outward Angles, together with the inward ones, make twice as many Right Angles as the Figure has Sides; but from the last Theorem, all the inward Angles, together, make twice as many Right Angles, abating four, as the Figure hath Sides. Wherefore the outward Angles are, all together, equal to four Right Angles. W. W. D. PROPOSITION XXXIII. THEORE M. Two Right Lines which join two equal and parallel Right Lines towards the fame Parts, are also equal and parallel. L ET the parallel and equal Right Lines A B, CD, be joined, towards the fame Parts, by the Right Lines AC, BD. I fay, A C, B D, are equal and parallel. For draw BC. Then, because A B is parallel to CD, and B C falls upon them, the alternate Angles ABC, BCD, are * equal. Again, because A B is equal to CD, and * 29 of this. BC is common; the two Sides A B, B C, are equal to the two Sides, B C, CD; but the Angle ABC is alfo equal to the Angle BCD; therefore the Bafe A C ist equal to the Base BD: And the Triangle A B C † 4 of this. equal to the Triangle BCD; and the remaining Angles equal to the remaining Angles, each to each, which fubtend the equal Sides. Wherefore the Angle ACB is equal to the Angle C B D. And becaule the Right Line BC, falling upon two Right Lines AC, BD, makes the alternate Angles AC B, C B D, equal between themselves; A C is parallel to B D. But it t 29 of this. has been proved alfo to be equal to it. Therefore, two Right Lines, which join two equal and parallel Right Lines, towards the fame Parts, are also equal and parallel; which was to be demonftrated. Defin. A Parallelogram is a Quadrilateral Figure, each of whofe oppofite Sides are parallel. PROPOSITION XXXIV. THEOREM. The oppofite Sides and oppofite Angles of any Pa• rallelogram are equal; and the Diameter divides the fame into two equal Parts. LETA BCD be a Parallelogram, whofe Diameter is B C. I fay, the oppofite Sides and oppofite Angles are equal between themselves, and the Diameter B C bifects the Parallelogram. For, because A B is parallel to C D, and the Right Line B C falls on them, the alternate Angles ABC, 29 of this. BCD, are equal between themselves; again, becaufe AC is parallel to B D, and B C falls upon them, the alternate Angles A C B, and C B D, are equal to one another. Wherefore A B C, CBD, are two Triangles, having two Angles ABC, BCA, of the one, equal to two Angles BCD, CBD, of the other, each to each; and likewife one Side of the one equal to one Side of the other, viz. the Side B C between the equal Angles, which is common. There26 of this, fore the remaining Sides fhall be + equal to the remaining Sides, each to each, and the remaining Angle to the remaining Angle. And fo the Side A B is equal to the Side CD, the Side A C to B D, and the Angle B A C to the Angle B D C. And because the Angle A B C is equal to the Angle B C D, and the Angle CBD to the Angle AC B; therefore the whole Angle ABD is equal to the whole Angle A CD: But it has been proved that the Angle BAC is alfo equal to the Angle BDC. Wherefore, the oppofite Sides and Angles of any Parallelogram are equal between themselves. I fay, moreover, that the Diameter bifects it. For because A B is equal to CD, and BC is common, the two Sides A B, BC, are each equal to the two Sides DC, CB; and the Angle A B C is alfo equal 4 of this. to the Angle D C B. Therefore the Base AC is ‡ equal |