PROBLEM XII. 14. To find the centre of a given circumference or of a given 15. Scholium. By the same construction a circumference may be made to pass through any three given points; or a circle circumscribed about a given triangle. PROBLEM XIII. 16. To inscribe a circle in a given triangle. Let A B C be the given triangle. A Bisect any two of its angles. With the point D, where the two bisecting lines meet, as a centre, with a radius equal to the distance of D from any one of the sides, describe a circle, and it will be the circle required. E F B G Draw the perpendicular DE, DF, DG. The angles at 4 are equal by construction, and the angles AED and AFD are each right angles; therefore the triangles ADE and AFD are equiangular (I. 35), and the side A D is common; therefore the triangles are equal (I. 41), and DE DF. In like manner DE DG. Therefore the circle described from D as a centre with the radius DE will pass through the points F and G; and since the angles at E, F, G are right angles, the sides of the triangle ABC are == tangents; therefore the circle EFG is inscribed in the triangle ABC (III. 20). 17. Scholium. The lines bisecting the angles of a triangle all meet in the same point. PROBLEM XIV. 18. Through a given point to draw a tangent to a given circumference. 1st. If the given point is in the circumference. Erect a perpendicular to the radius at the given point (3). 2d. If the given point is without the circumference. B Join the given point A with the centre C of the given circle BDE; on A Cas a diameter describe a circle cutting the given circle in B and E D A D. Draw A B and A D, and each will be tangent to the given circle through the given point. For drawing the radii C'B, CD, the angles B, D are each right angles (III. 23); therefore AB, AD are tangents to the given circle. 19. Corollary. The tangents A B, A D are equal (I. 50). PROBLEM XV. 20. Upon a given straight line to describe a segment of a circle which shall contain a given angle. CB, describe a circle AG BF; BFA is the segment required. As BD is perpendicular to the radius C B at B, it is a tangent to the circle, and hence the angle ABD is measured by half the arc A G B (III. 54); and any angle B F A inscribed in the segment BFA is also measured by half the arc AG B (III. 21), and is therefore equal to the angle ABD or the given angle. 21. Corollary. If the given angle is a right angle, the required segment would be a semicircle described on the given line as a diameter. PROBLEM XVI. 22. To divide a given line into parts proportional to given lines. From A cut off AD, DE, EF equal respectively to M, N, O. Join B to F, and through D and E draw lines parallel to BF. These parallels divide the line as required (II. 16). 23. Corollary. By taking M, N, O equal, the given line can be divided into equal parts. PROBLEM XVII. 24. To find a fourth proportional to three given lines. lines AF, AG. From AF cut off ABM, BC=N, and from AG cut off AD = 0. Join BD and through C draw CE parallel to BD; then DE is the required fourth proportional (II. 16). 25. Corollary. By taking A B equal to M, and AD and BC each equal to N, a third proportional can be found to M and N. PROBLEM XVIII. 26. To find a mean proportional between two given lines. Let it be required to find a mean MN D BD is the mean propor at B draw BD perpendicular to AC. 27. Definition. When a line is divided so that one segment is a mean proportional between the whole line and the other segment, it is said to be divided in extreme and mean ratio. PROBLEM XIX. 28. To divide a given line in extreme and mean ratio. Let it be required to divide A B in extreme and mean ratio. tre C and radius CB and produce AC to meet the circumference in F; then A F is a secant and A B a tangent of the circle DFB, and therefore (III. 64) AB: AB-AB-AD: AD AB 2 CB=DF AF-AB-AF-DF-AD-AE and the proportion becomes or (Pn. 16) AE: ABEBAE AB:AEAE: EB PROBLEM XX. 29. Through a given point in a given angle to draw a line so that the segments included between the point and the sides of the angle may be in a given ratio. Let it be required to draw through "the point D within the angle B a line so that AD:DC=M: N. Draw DE parallel to A B. B MN E A D and produce CD to A, and AC is the line required (II. 16). PROBLEM XXI. 30. The base, an adjacent angle, and the altitude of a triangle given, to construct the triangle. At A of the base A B draw an indefinite line AC making the angle A equal to the given angle; at any point in AB, as D, draw the perpendicular DE equal E G C D B to the given altitude; through E draw EF parallel to A B cutting A C in G; join G B, and A G B is the triangle required. |