which meets it, the line which meets shall touch the circle. Let any point D be taken without the circle ABC, and from it let two straight lines DCA and D B be drawn, of which D C A cuts the circle, and DB meets it; if the rectangle A D, DC be equal to the square of DB; DB touches the circle. Draw (III. 17.) the straight line D E touching the circle A B C, finds its centre F, and join FE, FB, FD; and then FED is a right angle (III. 18.): And because DE touches the circle A B C and D C A cuts it, the rectangle AD, DC is equal (III. 36.) to the square of DE: But the rectangle A D, DC is, by hypothesis, equal to the square of DB: Therefore the square of DE is equal to the square of DB; and the straight line D E equal to the straight line DB: And F E is equal to F B, wherefore D E, EF D E are equal to DB, BF, each to each; and the base FD is common to the two triangles D E F, DBF; therefore the angle D E F is equal (1. 8.) to the angle DBF; B but D E F is a right angle, therefore also DB F is a right angle: And FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches (III. 16.) the circle: Therefore DB touches the circle AB C. A Wherefore, if from a point, &c. Q. E. D. GEOMETRICAL EXERCISES. EXERCISES ON BOOK I. PROBLEMS. No. I. 1. To bisect a right angle; that is, to divide it into three equal parts. Let KBC be a right angle: It is required to trisect it. In BK take any point a: on BA describe (Euclid 1. 1.) the equilateral triangle ABD; and bisect (E. 1. 9.) the angle ABD by the straight line BE; then the angles CBD, DBE, and EBA are equal to one another. For the triangle ABD being equilateral the angle ABD (E. I. 5. and 32.) is one third of two right angles, or two thirds of the right angle ABC; therefore the angle CBD is one third of the right angle CBA; but since the angle DBA is bisected by BE, the angles DBE and E BA are each of them equal to one third of a right angle; wherefore the angles CBD, DBE, and EBA are equal to one another. 2. To trisect a given straight line. Let AB be the given straight line: It is required to divide it into three equal parts. Upon A B describe (E. 1. 1.) the equilateral triangle ABC; bisect (E. 1. 9.) the angles CAB and CBA by the lines A D and BD meeting in D; and from D draw (E. I. 31.) DE A E C parallel to CA, and DF parallel to CB: Then AB is trisected in the points E and F. For since DE is parallel to CA and AD meets them, therefore (E. 1. 29.) the angles DAC and ADE are equal; but (by constr.) the angle EAD is equal to the angle DAC; therefore the angle ADE is equal to the angle EAD; wherefore (E. 1. 6.) AE is equal to ED. For the same reason FB is equal to FD. But DE being parallel to CA and DF to CB, the angle DEF is equal to the angle CAB, and DFE to CBA; and therefore the remaining angle EDF is equal to the remaining angle ACB; hence the triangle EDF is equiangular, and therefore ED, EF, FD are equal to one another; but ED has been shown to be equal to AE, and FD to FB, therefore AE, EF, FB` are equal to one another. 3. Through a given point to draw a straight line which shall make equal angles with two straight lines given in position. B E Let P be the given point, and BE and CF the lines given in position. It is required to draw, through the point P, a straight line which shall make equal angles with BE and CF. Produce BE and CF to meet in A; bisect (E. 1. C D A P F 9.) the angle CAB by the line AD; and from P let fall (E. 1. 12.) the perpendicular PD, and produce it on both sides to E and F. Then FE is the line required. For, by construction, the angle DAE is equal to the angle DAF, and the angles at D are right angles, and moreover AD is common to both the triangles L ADE and ADF, therefore (E. 1. 26.) the angle AED is equal to the angle AFD. 4. From two given points on the same side of a line given in position, to draw two lines which shall meet in that line, and make equal angles with it. Let C and D be the two given points, and AH the line given in position. From c let fall (E. 1. 12.) the perpendicular CG, and produce it to F making GF equal to GC; join FD, CE; then CE and ED are the lines required. Since CG is equal to GF, and GE is common to the two triangles CGE and FGE, and the angles at G are right A D E angles, therefore (E. 1. 4.) the triangles CGE and FGE are equal, and the angle CEG is equal to the angle FEG; but (E. I. 15.) the angle DEH is equal to the angle FEG, therefore the angle CEG is equal to the angle DEH. 5. On a given line to describe an isosceles triangle, whose perpendicular height is equal to the base. 6. To make an angle equal to half a right angle. 7. Given the diagonal of a square to construct it. 8. Given the base, the perpendicular height, and one of the angles at the base of a triangle, to construct it. 9. Trisect two right angles. 10. Divide a right angle into six equal parts. 11. Divide a right angle into four equal parts. 12. Given the hypotenuse of a right-angled triangle, and the difference of the two acute angles, to construct the triangle. 13. To construct a right-angled triangle, having given the hypotenuse, and one of the acute angles equal to one third a right angle. Show that the side opposite to the given angle is equal to one half the hypotenuse. 14. Given the two sides, and an angle opposite to one of the given sides of a triangle, to construct it. Show that, in general, there are two triangles answering the conditions of the problem. 15. Given the base, one of the other sides, and the perpendicular on the base of a triangle, to construct it. 16. On a given base, to construct an isosceles triangle having the angle at the vertex equal to one third a right angle. 17. From the vertex of a given triangle, to draw to the base a straight line which shall be less than the greater side by a given line. 18. Given the perpendicular and the equal side of an isosceles triangle, to construct it. 19. Through a given point P, to draw a straight line, which shall cut off equal parts from two straight lines AB and AC, cutting one another in a. 20. Given the vertical angle and the perpendicular height of an isosceles triangle, to construct it. 21. Given the base, the less angle at the base, and the difference of the sides of a triangle, to construct it. 22. From a given point, to draw the shortest line possible to a given straight line. 23. In a straight line given in position, but indefinite in length, to find a point, which shall be equidistant from each of two given points, on the same side of the given line, and in the same plane with it. 24. To draw a straight line from a given point to |