a right angle; therefore each of the angles CEA, EAC is half a right angle for the same reason, each of the angles CEB, EBC is half a right angle; therefore AEB is a right angle; and because EBC is half a right angle, DBG is also (3. 1.) half a right angle, for they are vertically opposite; but BDG is a right angle, because it is equal to the corresponding 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 (9. 2.) 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 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 (17. 2.) 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 (17. 2.) 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 (17.2.) to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD: but the squares of AD, DG are equal (17. 2.) 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. XXIV. PROB. To divide a given straight line into two parts, so that the rectangle con tained 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 (16. 2.) the square ABDC; bisect (8. 1.) AC in E, and join BE; produce CA to F, and make EF equal to EB; and upon AF describe (16.2.) the square FGHA; AB is divided in H, so that the rectangle AB, BH is equal to the square of 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 AE, is equal (22. 2.) to the square of EF: but EF is equal to EB; therefore the rectangle CF, FA, together with the square of G H B AE, is equal to the square of EB: and the squares of BA, AE are equal (17. 2.) 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 PROP. XXV. THEOR. K D If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC are equal to twice the rectangle AB, BC, together with the square of AC. gno- A H C B G K Upon AB describe (16. 2.) the square ADEB, and construct the figure as in the preceding propositions: and because AG is equal (19. 2.) to GE, add to each of them CK; the whole AK is therefore equal to the whole CE; therefore AK, CE are double of AK: but AK, CE are the mon AKF, together with the square CK; therefore the gnomon AKF, together with the square CK, is double of KA: but twice the rectangle AB, BC is double of AK, for BK is equal (Cor. 20. 2.) to BC: therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle AB, BC: to each of these equals add HF, which is equal to the square AC: therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC, and the square of AC: but the gnomon AKF, together with the of D FE squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC: therefore the squares of AB and BC are equal to twice the rectangle AB, BC, together with the square AC. Wherefore, if a straight line, &c. Q. E. D. PROP. XXVI. 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 (10. 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 (20. 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 (17. 2.) to the squares of BD, DA, because the angle at D is a right angle; and the square of CA 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 BC, CD. Therefore, in obtuse angled triangles, &c. Q. E. D. B PROP. XXVIII. THEOR. C D 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 (10. 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 (25. 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, ᎠᎪ are equal to twice the rectangle CB BD, and the squares of AD, DC: but the square of AB is equal (17. 2.) to the 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 twice the rectangle CB, BD. B B D C D Secondly, Let AD fall without the triangle ABC: then, because the angle at D is a right angle, the angle ACB is greater (5. 2.) than a right angle; and therefore the square of AB is equal (26. 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 BC, and twice the rectangle BC, CD: but because BD is divided into two parts in C, the rectangle DB, BC is composed of 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 squares of AB, BC are equal (17. 2.) to the square of AC and twice the square of BC. Therefore, in every triangle, &c. Q. E. D. A THE ELEMENT OF GEOMETRY. BOOK III. DEFINITIONS. I. WHEN a straight line meets a circle, and being produced does not cut the circle, it may be said to touch the circle, and may be called a tangent. II. When the perpendiculars drawn to straight lines from the centre of a circle are equal, the straight lines may be said to be equally distant from the centre. III. And the straight line on which the greater perpendicular falls, may be said to be farther from the centre. IV. The part of a circle contained by a straight line and the circumference which it cuts off, may be called a segment of a circle, and the straight line may be called the base of the segment. |