At A, B draw AC, BD perpendicular to AB (1. 11); bisect the angles CAB, ABD by the straight lines AE, BE (1.9); bisect the angle ABE by the straight line BF, meeting AE in F; through F draw FG parallel to EB (1. 31). Then AB shall be divided in the point G, so that the square on AG is double the square on GB. Proof. The angles EAB, EBA are each equal to half a right angle (con.), therefore the angle AEB is a right angle (1. 32); and the angle AFG is a right angle, being equal to the angle AEB (1. 29). The angle AFG is a right angle, and the angle FAG is half a right angle, therefore the angle FGA is half a right angle (1. 32); wherefore FA is equal to FG (1. 6). Again, the angle GFB is equal to the alternate angle EBF (1. 29), and the angle GBF is equal to the angle EBF (con.), therefore the angle GFB is equal to the angle GBF (ax. 1); wherefore GF is equal to GB: but FA has been proved to be equal to FG; therefore FA, FG, GB are equal to one another. The square on AG is equal to the squares on AF, FG (1. 47); therefore the square on AG is double the square on GB. Wherefore the straight line AB is divided in the point G, so that the square on AG is double the square on GB. Q.E.F. 4. In any triangle BAC, a line AD is drawn, bisecting BC in D. Show that the sum of the squares on AB, AC is equal to twice the sum of the squares on AD, BD. Draw AE perpendicular to BC (1. 12). Proof. In the obtuse-angled triangle ADB, the square on AB is greater than the squares on AD, DB, by twice the rectangle BD, DE (II. 12); and in the acute-angled triangle ADC, the square on AC is less than the squares on AD, DC, by twice the rectangle CD, DE (II. 13); but BD is equal to DC (hyp.); wherefore the squares on AB, AC are double of the squares on AD, BD. Q.E.D. 5. Show that if a straight line be divided into two parts, so that the rectangle contained by the whole and the first part is equal to the square of the second part, then must the squares on the whole line and on the first part be equal to three times the square on the second part. (See deduction to II. 11.) 6. From AC, the diagonal of a square ABCD, cut off AE equal to one-fourth of AC, and join BE, DE. Show that the figure BADE equals twice the square on AE. E Proof-Because AE is one-fourth of AC (hyp.), therefore the triangle AED is one-fourth of the triangle ADC; similarly, the triangle AEB is one-fourth of the triangle ABC; therefore the figure BADE is one-fourth of the square ABCD. EUCLID, BOOK II. DEFINITIONS AND GENERAL QUESTIONS. Enunciate the proposition which proves that the area of a triangle is represented by half the rectangle which has the same base and altitude. 'If a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram shall be double of the triangle' (1. 41). 2. Deduce from this that a square on a straight line is equal to four times the square on half the line. Let the straight line AB be bisected in the point C. Then the square on AB shall be equal to four times the square on AC or BC. Proof. The square on AB is equal to the rectangles contained by AB, AC, and AB, BC (II. 2); but AB is equal to twice AC or BC (hyp.); therefore the square on AB is equal to twice the square on AC and twice the square on BC: wherefore the square on AB is equal to four times the square on AC or BC. Q.E.D. 3. Divide a given line so that the rectangle contained by the whole line and one part may be twice the square on the other part. The square on BE is equal to the squares on BG, GE (1. 47): and because the straight line CF is bisected in G, and produced to B, therefore the rectangle CB, BF, together with the square on GF, is equal to the square on GB (11. 6); but BF is equal to DE (1. 34), and GF has been proved to be equal to GC; therefore the square on BG is equal to the rectangle BC, DE, together with the square on GC:. wherefore the square on BE is equal to the squares on EG, GC, together with the rectangle BC, DE; but the square on CE is equal to the squares on EG, GC (1. 47) ; therefore the square on BE is equal to the square on CE, together with the rectangle BC, DE. Q.E.D. 8. Enunciate (only) Props. 12 and 13 of Second Book. The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals. Let ABCD be a parallelogram. Then the sum of the squares on the diagonals AC, BD shall be equal to the sum of the squares on AB, BC, CD, DA, that is, twice the squares on AB, BC. A BE Draw AE, DF perpendicular to BC, and BC produced (1. 12). Proof. In the triangles ABE, DCF, the angle AEB is equal to the angle DFC (ax. 11), the angle ABE is equal to the angle DCF (1. 29), and the side AB is equal to the side DC (1. 34); therefore the side BE is equal to the side CF (1. 26). The square on BD is greater than the squares on BC, CD, by twice the rectangle BC, CF (11. 12); and the square on AC is less than the squares on AB, BC, by twice the rectangle CB, BE (II. 13); but the rectangle BC, CF is equal to the rectangle CB, BE; wherefore the squares on AC, BD are equal to the squares on AB, BC, CD, DA, that is, twice the squares on AB, BC. Q.E.D. 9. If the square on the perpendicular from the vertex of a triangle to the base is equal to the rectangle contained by the segments of the base, the vertical angle is a right angle. Let ABC be a triangle, such that the square on the perpendicular AD is equal to the rectangle BD, DC. Then the angle BAC shall be a right angle. B Proof-The square on AB is equal to the squares on BD, AD (1. 47); and the square on AC is equal to the squares on DC, AD; therefore the squares on AB, AC are equal to the squares on BD, DC, and twice the square on AD (ax. 2); but the square on AD is equal to the rectangle BD, DC (hyp.); therefore the squares on AB, AC are equal to the squares on BD, DC, and twice the rectangle BD, DC; |