PROBLEM II. 2. From a given point without a straight line to draw a perpendicular to that line. Let C be the point and A B the line. C cutting A B in two points E and F; with PROBLEM III. 3. From a given point in a straight line to erect a perpendicular to that line. Let C be the given point and A B the given line. F With Cas a centre describe an arc cutting A B in D and E; with D and E as centres, with a radius greater than 4DC, describe arcs intersecting at F. Draw CF, and it is the perpendicular required (converse of -B C D E I. 53). Second Method. With Cas a centre describe an arc DEF; take the distances DE and EF equal to CD, and from E and Fas centres, with a radius greater than half the distance from E to F, describe arcs intersecting at G. Draw CG, and it is the perpendicular required (III. 33; III. 16; III. 15). Third Method. With any point, D, without the line A B, with a radius equal to the distance from D to C, describe an arc cutting A B at E; draw the diameter EDF. Draw CF, and it is the perpendicular required (III. 23). PROBLEM IV. 4. To bisect a given arc, or angle. 1st. Let A B be the given arc. Draw the chord AB and bisect it with a perpendicular (1; III. 16). 2d. Let C be the given angle. With Cas a centre describe an arc cutting the sides of the angle in A and B; bisect the arc A B with the line CD, and it will also bisect the angle C (III. 11). D PROBLEM V. 5. At a given point in a straight line to make an angle equal to a given angle. Let A be the given point in the line AB, and C the given angle. With C as a centre describe an arc DE cutting the sides of the angle C; with as a centre, with the same radius, describe an arc; with Fas a centre, with a radius equal to the distance from D to E, describe an arc cutting the A A G F E D arc FG. Draw AG. The angle A = C (III. 12; III. 11). PROBLEM VI. 6. Through a given point to draw a line parallel to a given straight line. 7. Two angles of a triangle given, to find the third. 8. The three sides of a triangle given, to construct the triangle. C Take A B equal to one of the given sides; with A as a centre, with a radius equal to another of the given sides, describe an arc, and with B as a centre, with a radius equal A B to the remaining side, describe an arc intersecting the first are at C. Draw AC and CB, and ACB is evidently the triangle required. PROBLEM IX. 9. Two sides and the included angle of a triangle given, to construct the triangle. Draw A B equal to one of the given sides; at B make the angle ABC equal to the given angle (5), and take BC equal to the other given side; join A and C, and ABC is evidently the triangle required. C A B PROBLEM X. 10. Two angles and a side of a triangle given, to construct the triangle. If the angles given are not both adjacent to the given side, find the third angle by (7). Then draw A B equal to the given side, and at B make an angle ABC equal to one of the angles adjacent to A B, and at A make an angle BAC equal to the other angle adjacent to AB, and ABC is evidently the triangle required. C A B PROBLEM XI. 11. Two sides of a triangle and the angle opposite one of them Bas a centre, with a radius equal to the other given side, describe an arc cutting A C. If the given angle A is acute, 1st. The given side BC, opposite the given angle, may be less than the other given side; then the arc described from Bas a centre B A D E C will cut AC in two points, C and D, 2d. The given side opposite the given angle may be equal to the perpendicular BE; then the arc described from Basa centre will touch AC, and the right triangle A BE is the only one that can satisfy the given conditions. 3d. The side opposite the given angle may be greater than the other given side; then the arc described from B as a centre AC will cut A Bin C, and in another point on the other side of A. In this case there can be but one triangle ABC satisfying the given conditions, the triangle formed on the opposite side of A B containing not the given angle but its supplement. 4th. If the given angle is obtuse, the given side opposite the given angle must be greater than the other given side, and as in the last case above there can be but one solution. 12. Scholium. If the side opposite the given angle A is less than the perpendicular, or if the given angle is right or obtuse, and at the same time the side opposite the given angle is less than the other given side, the solution is impossible. 13. Corollary. From this and the preceding Problem and Theorems VIII., IX., and XIV. of Book I., it follows that with the exception of the ambiguity pointed out in the first part of this Problem, two triangles are equal if any three parts, of which one is a side, of the one are equal to the corresponding parts of the other. |