Then KHM is a straight line; and FGL is a straight line. ... KFLM a parallelogram. And the figure ABCD is equal to it. And because at the point H, in the straight line GH, the two straight lines KH, HM, on the opposite sides of it, make the adjacent angles together equal to two right angles, Therefore KH is in the same straight line with HM (I. 14). And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal (I. 29). Add to each of these equals the angle HGL; Therefore the angles MHG, HGL are equal to the angles HGF, HGL (Ax. 2). But the angles MHG, HGL are equal to two right angles (I. 29); Therefore also the angles HGF, HGL are equal to two right angles, And therefore FG is in the same straight line with GL (I. 14). And because KF is parallel to HG, and HG parallel to ML (Const.); Therefore KF is parallel to ML (I. 30). And KM, FL are parallels (Const); Therefore KFLM is a parallelogram (Def. 35). And because the triangle ABD is equal to the parallelogram HF, and the triangle DBC equal to the parallelogram GM (Const.), Therefore the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM (Ax. 2). Therefore, the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E. Q. E. F. COROLLARY.-From this it is manifest how to apply to a given straight line a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure-namely, by applying to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle; and so on (I. 44). Proposition 46.-Problem. To describe a square upon a given straight line. It is required to describe a square upon AB. CONSTRUCTION.-From the point A draw AC at right angles to AB (I. 11), And make AD equal to AB (I. 3). Through the point D draw DE parallel to AB (I. 31). Through the point B draw BE parallel to AD (I. 31). Then ADEB shall be the square required. PROOF.-Because DE is parallel to D AB, and BE parallel to AD (Const.), A Therefore AB is equal to DE, and AD to BE (I. 34). Therefore the four straight lines BA, AD, DE, EB are equal to one another (Ax. 1), And the parallelogram ADEB is therefore equilateral. For since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are together equal to two right angles (I. 29). But BAD is a right angle (Const.); Therefore also ADE is a right angle (Ax. 3). But the opposite angles of parallelograms are equal (I. 34); Therefore each of the opposite angles ABE, BED is a right angle (Ax. 1); It is equilateral. tangular. Therefore the figure ADEB is rectangular; and it has It is recbeen proved to be equilateral; therefore it is a square (Def. 30). Therefore, the figure ADEB is a square, and it is described ..a square. upon the given straight line AB. Q. E. F COROLLARY.-Hence every parallelogram that has one right angle has all its angles right angles. Proposition 47.-Theorem. In any right-angled triangle, the square which is described upon the side opposite to the right angle is equal to the squares described upon the sides which contain the right angle. Let ABC be a right-angled triangle, having the right PROOF.-Because the angle BAC is a right angle (Hyp.), and that the angle BAG is also a right angle (Def. 30), The two straight lines AC, AG, upon opposite sides of AB, make with it at the point A the adjacent angles equal CAG is a to two right angles; straight A ABD = Therefore CA is in the same straight line with AG (I. 14). For the same reason, AB and AH are in the same straight line. Now the angle DBC is equal to the angle FBA, for each of them is a right angle (Ax. 11); add to each the angle ABC. Therefore the whole angle DBA is equal to the whole angle FBC (Ax. 2). And because the two sides AB, BD are equal to the two sides FB, BC, each to each (Def. 30), and the angle DBA equal to the angle FBC; Therefore the base AD is equal to the base FC, and the triangle ABD to the triangle FBC (I. 4). Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and between the same parallels BD, AL (I. 41). And the square GB is double of the triangle FBC, because they are on the same base FB, and between the same parallels parallelo- FB, GC (I. 41). Hence gram BL =square GB, and But the doubles of equals are equal (Ax. 6), therefore the parallelogram BL is equal to the parallelo gram CL = square HC. square GB. In the same manner, by joining AE, BK, it can be shown that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB, HC (Ax. 2); And the square BDEC is described on the straight line BC, and the squares GB, HC upon BA, AC. Therefore the square described upon the side BC is equal to the squares described upon the sides BA, AC. Therefore, in any right-angled triangle, &c. Q. E. D. Proposition 48.-Theorem. If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle. Let the square described upon BC, one of the sides of the triangle ABC, be equal to the squares described upon the other sides BA, AC; .. BC2 = BA2+AC2 Draw AD at The angle BAC shall be a right angle. angles to AC (I. 11). Make AD equal to BA (I. 3), and join DC. is equal to the square on BA. To each of these add the square on AC. Therefore the squares on DA, AC are equal to the squares on BA, AC (Ax. 2). But because the angle DAC is a right angle (Const.), the square on DC is equal to the squares on DA, AC (I. 47), And the square on BC is equal to the 13 squares on BA, AC (Hyp.); angles to AC. (Do not produce BA.) Then Therefore the square on DC is equal to the square on BC DC2 = (Ax. 1); And therefore the side DC is equal to the side BC. And because the side DA is equal to AB (Const.), and AC common to the two triangles DAC, BAC, the two sides DA, AC are equal to the two sides BA, AC, each to euch. And the base DC has been proved equal to the base BC; BC2, and DC BC. Hence Therefore the angle DAC is equal to BAC (I. 8). Therefore also BAC is a right angle (Ax. 1). EXERCISES. PROP. 1-15. 1. From the greater of two given straight lines to cut off a portion which is three times as long as the less. 2. The line bisecting the vertical angle of an isosceles triangle also bisects the base. 3. Prove Euc. I. 5, by the method of super-position. 4. In the figure to Euc. I. 5, show that the line joining A with the point of intersection of BG and FC, makes equal angles with AB and AC. 5. ABC is an isosceles triangle, whose base is BC, and AD is perpendicular to BC; every point in AD is equally distant from B and C. 6. Show that the sum of the sum and difference of two given straight lines is twice the greater, and that the difference of the sum and difference is twice the less. 7. Prove the same property with regard to angles. 8. Make an angle which shall be three-fourths of a right angle. 9. If, with the extremities of a given line as centres, circles be drawn intersecting in two points, the line joining the points of intersection will be perpendicular to the given line, and will also bisect it. 10. Find a point which is at a given distance from a given point and from a given line. 11. Show that the sum of the angles round a given point are together equal to four right angles. 12. If the exterior angle of a triangle and its adjacent interior angle be bisected, the bisecting lines will be at right angles. 13. If three points, A, B, C, be taken not in the same straight line, and AB and AC be joined and bisected by perpendiculars which meet in D, show that DA, DB, DC are equal to each other. PROP. 16-32. 14. The perpendiculars from the angular points upon the opposite sides of a triangle meet in a point. 15. To construct an isosceles triangle on a given base, the sides being each of them double the given base. |