E F (15. 1.) half a right angle, for they are vertically opposite; but BDG is a right angle, because it is equal (29. 1.) to the alternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side BD is equal (6. 1.) to the side DG: again, because EGF is half a right angle, and that the angle at F is a right angle, because it is equal (34. 1.) to the opposite angle ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF; wherefore also the side GF is equal (6. 1.) to the side FE. And because EC is A B D G equal to CA, the square of EC is equal to the square of CA; therefore the squares of EC, CA are double of the square of CA: but the square of EA is equal (47. 1.) to the squares of EC, CA; therefore the square of EA is double of the square of AC: again, because GF is equal to FE, the square of GF is equal to the square of FE: and therefore the squares of GF, FE are double of the square of EF: but the square of EG is equal (47. 1.) to the squares of GF, FE; therefore the square of EG is double of the square of EF and EF is equal to CD; wherefore the square of EG is double of the square of CD: but it was demonstrated, that the square of EA is double of the square of AC; therefore the squares of AE, EG are double of the squares of AC, CD: and the square of AG is equal (47. 1.) to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD: but the squares of AD, GD are equal (47. 1.) to the square of AG; therefore the squares of AD, DG are double of the squares of AC, CD: but DG is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD. Wherefore if a straight line, &c. Q. E. D. PROP. XI. PROB. To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part. Let AB be the given straight line: it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part. Upon AB describe (46. 1.) the square ABDC; bisect (10. 1.) AC in E, and join BE; produce CA to F, and make (3. 1.) EF equal to EB, and upon AF describe (46. 1.) the square FGHA; AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. A F G H B 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 AE, is equal (6. 2.) to the square of EF: but EF is equal to EB; therefore the rectangle CF, FA, together with the square of AE, is equal to the square of EB: and the squares of BA, AE are equal (47. 1.) to the square of EB, because the angle 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 AE, which is common to both, therefore the remaining rectangle CF, FA is equal to the square of AB; and the 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 remainder 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. E C. K D 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 obtuse angle. Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn (12. 1.) perpendicular to BC produced: the square of AB is greater than the squares of AC, CB by twice the rectangle BC, CD. A Because the straight line BD is divided into two parts in the point C, the square of BD is equal (4. 2.) to the squares of BC, CD, and twice the rectangle BC, CD: 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 (47. 1.) to the squares of BD, DA, because the angle at B C D D is a right angle; and the square of CA is equal (47. 1.) 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 BC, CD. Therefore, in obtuse angled triangles, &c. Q. E. Ď. PROP. XIII. THEOR. IN 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 contained 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.* 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 (12. 1.) 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. A First, Let AD fall within the triangle ABC; and because the straight line CD is divided into two parts in the point D, the squares of CB, BD are equal (7. 2.) 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, BD, DA are equal to twice the rectangle CB, BD, and the squares of AD, DC: but the square of B AB is equal (47. 1.) to the squares of D A 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 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 (16. 1.) than a right angle; and therefore the square of AB is equal (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 AC, and twice the square of B * See Note. C D BC, and twice the rectangle BC, CD: but because BD is divid-" ed into two parts in C, the rectangle DB, BC is equal (3. 2.) to the rectangle BC, CD and 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 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 square of AB, BC are equal (47. 1.) to the square of AC and twice the square of BC. fore, in every triangle, &c. Q. E. D. There A B C PROP. XIV. PROB. To describe a square that shall be equal to a given rectilineal figure.* Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A. Describe (45. 1.) the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it BE, ED are equal to one another, it is a square, and what was required is now done: but if they are not equal, produce one of them BE to F, and H 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 (5. 2.) 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 • See Note. |