b 10. 1. Upon A B describe a the square ABDC; bisect AC in E, Book II. and join BE; produce CA to F, and make " EF equal to EB, and upon AF describe a the square FGHA; AB is divided : 46. !. in H, so that the rectangle AB, BH is equal to the square of c 3. 1. AH. Produce GH to K: because the straight line AC is bisected in E, and produced to the point F, the rectangle CF, FA, together with the square of A E, is equal d to the square of EF: 6. 2. But EF is equal to EB; therefore the rectangle CF, FA, together with the square of A E, is equal to the square of EB : And the squares of BA, AE are equal e e 47. 1. to the square of EB, because the angle F G EAB is a right angle; therefore the rectangle CF, FA, together with the square of AE, is equal to the squares of BA, AE: Take away the square of A HB AE, which is common to both, therefore the remaining rectangle CF, FA, is equal to the square of AB; and the E figure FK is the rectangle contained by CF, FA, for AF is equal to FG; and AD is the square of AB; therefore FK is equal to AD: Take away the common part AK, and the re K D mainder FH is equal to the remainder HD: and HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the square of AH. Therefore the rectangle AB, BH is equal to the square of AH: Wherefore, the straight line AB is divided in H, so that the rectangle AB, BH, is equal to the square of AH. Which was to be done. PROP. XII. THEOR. In obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obluse angle. Book II. a 12. 1. b 4. 2. Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn a perpendicular to BC produced: The square of AB is greater than the squares of AC, CB, by twice the rectangle BC, CD. Because the straight line BD is divided into two parts in the point C, the square of BD is equal to the squares of BC, CD, and twice the rectangle BC, CD: A To each of these equals add the square of DA; and the squares of BD, DA, are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD: But the square of BA is equal to the squares of BD, DA, because the angle at D'is a right B С D angle; and the square of CĂ is equal to the squares of CD, DA: Therefore the square of BA is equal to the squares of BC, CA, and twice the rectangle BC, CD; that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BČ, CD. Therefore, “ in obtuse angled triangles,” &c. Q. E. D. c 47. 1. с PROP. XIII. THEOR. See N. In every triangle, the square of the side subtending any of the acute angles, is less than the squares of the sides containing that angle by twice the rectangle con. tained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. a 12. 1. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular a AD from the opposite angle : The square of AC, opposite to the angle B, is less than the squares of CB, BA, by twice the rectangle CB, BD. First, Let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D, the squares of CB, BD are equal to twice the rectangle contained by CB, BD, and the square of DC: To each of these equals add the square of AD; therefore the squares of CB, b 7. 2 d 16. 1. BD, DA are equal to twice the rectangle CB, BD, and the Book II. squares of AD, DC: But the square of AB is equal to the A c 47. 1. squares of BD, DA, because the angle BDA is a right angle; and the square of AC is equal to the squares of AD, DC: Therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD, that is, the square of AC alone is less than the squares of CB, BA, by B D twice the rectangle CB, BD. Secondly, Let AD fall without the triangle ABC: Then, because the angle at D is a right angle, the angle ACB is greater d than a right angle; and therefore the square of AB is equale e 12. 2. to the squares of AC, CB, and twice the rectangle BC, CD: To these equals add the square of BC, and the squares of AB, BC are equal to the square of B C D AC, and twice the square of BC, and twice the rectangle BC, CD: But because BD is divided into two parts in C, the rectangle DB, BC is equal to the rectangle BC, CD and f 3. 2. the square of BC: And the doubles of these are equal : Therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: Therefore the square of AC alone is less than the squares of AB, BC, А by twice the rectangle DB, BC. Lastly, Let the side AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifest that the squares of AB, BC are equal to the square of AC g 47. 1. and twice the square of BC: Therefore, “ in every triangle,” &c. Q. E. D. B С Book II. PROP. XIV. PROB. See N. To describe a square that shall be equal to a given rectilineal figure. a 15. 1. Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A. Describe a the rectangular parallelogram BCDE equal to H Н А B F equal, produce G E one of them BE C D to F, and make EF equal to ED, and bisect BF in G: and from the centre G, at the distance GB, or GF, describe the semicircle BHF; and produce DE to H, and join GH: Therefore because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the square of EG, is equal to the square of GF: But GF is equal to GH; therefore the rectangle BE, EF, together with the square of EG, is equal to the square of GH: But the squares of HE, EG are equal e to the square of GH: Therefore the rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EG: take away the square of EG, which is common to both ; and the remaining rectangle BE, EF is equal to the square of EH: but the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH: but BD is equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. Which was to be done. b 5. 2. c 47. 1. I. Equal circles are those of which the diameters are equal, or Book III. from the centres of which the straight lines to the circumfe. rences are equal. • This is not a definition but a theorem, the truth of which 6 is evident; for if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, • since the straight lines from their centres are equal.' II. a circle, when it meets the III. another, which meet, but IV. tant from the centre of a circle, when V. greater perpendicular falls, is said to |