each pair of opposite triangles is half the area of the parallelogram. THEOREM 4. If two equal triangles on the same base be cut by A a line parallel to the base, equal areas will be cut off from them. C F B D E W THEOREM 5. Lines drawn through the point of contact of two circles, and terminated by the circles, form four chords which are proportional, and the lines through the ends of which are parallel. A THEOREM 6. If a circle be described touching two parallels, and from the points of tangency secants be drawn intersecting the circle in the same point, and terminated by the opposite parallel, the diameter of the circle will be a mean proportional between the segments of the parallels. B Hypothesis. The lines PB and QA intersect at R, a point of the circle. Corollary. The same thing being supposed, prove THEOREM 7. If through any ver- G tex of a parallelogram a line be drawn meeting the two opposite sides produced without the parallelogram, the rectangle of the produced portion of such sides is equal to the rectangle of the sides of the parallelogram. Hypothesis. GD. BF = AB. BC. D THEOREM 8. If two triangles have one angle of the one. equal to one angle of the other, and the perpendiculars from the other two angles upon the opposite sides proportional, they are similar. Hypothesis. Angle C = angle C'. AP: BQ: A'P' : B'Q'. Conclusion. The triangles are similar. THEOREM 9. If the four sides of an inscribed quadrilateral taken consecutively form a proportion, the diagonal having the means on one side and the extremes on the other side divides it into two triangles of equal area. (Book IV., Ex. Th. 9.) THEOREM 10. The rectangle of two sides of a triangle is equal to the rectangle of its altitude above the third side and the diameter of the circumscribed circle. THEOREM 11. The area of a triangle is equal to the product of the three sides divided by twice the diameter of the circumscribed circle. THEOREM 12. If two parallel tangents to a circle are intercepted by a third tangent, the rectangle of the segments of the latter is equal to the square of the radius of the circle. THEOREM 13. If a chord be drawn parallel to the tangent at the vertex of an inscribed triangle, the portion of the triangle cut off by the chord is similar to the original triangle. THEOREM 14. If from the middle point A of the arc subtended by a fixed chord a second chord be drawn intersecting the fixed one, the rectangle contained by the whole of that second chord and the part of it intercepted between the fixed chord and the point A is a constant whatever be the direction of the second chord. Show to what square or area the constant area is equal. THEOREM 15. If in two triangles any angle of the one is equal to some angle of the other, their areas are to each other as the rectangles of the sides which contain the equal angles. THEOREM 16. If two chords of a circle intersect each other at right angles, the sum of the squares of the four segments is equal to the square of the diameter. THEOREM 17. If on each of the sides of an angle having its vertex at O two points A and B on the one side and P and on the other be taken such that OP OA :: OB: 0Q, the four points A, B, P, and Q will lie on a circle (§§ 244, 392). THEOREM 18. If at any point outside P A B of two circles a point be chosen from which the tangents to the two circles shall be equal in length, and from this point secants to each circle be drawn, the four points of intersection will lie on a third circle. Apply 429, 4, to the case of each circle. THEOREM 19. Conversely, if two circles be intersected by a third, and secants be drawn through each pair of points of intersection, the tangents to the circles from the point of intersection of the secants will be equal in length. THEOREM 20. If the common secant of two intersecting circles be drawn, the tangents to the two circles from each point of this secant will be equal in length. NUMERICAL EXERCISES. 1. If one of two similar triangles has its sides 50 per cent longer than the homologous sides of the other, what is the ratio of their areas? 2. The owner of a rectangular farm containing 10,000 square yards finds that it measures 5 inches x 20 inches on a map. What are its length and breadth? BOOK VI. REGULAR POLYGONS AND THE CIRCLE. CHAPTER I. PROPERTIES OF INSCRIBED AND CIRCUMSCRIBED REGULAR POLYGONS. THEOREM I. 44%. If a circle be divided into any number of equal arcs, and a chord be drawn in each arc, these chords will form a regular polygon. Hypothesis. A, B, C, D, E, equidistant points around a circle, separating it into equal arcs; AB, BC, etc., the chords of those arcs. Conclusion. The polygon ABCDE is regular (§ 152). Proof. I. Because the sides are by hypothesis all chords of equal arcs, they are all equal to each other. II. Take any two angles of the poly B E D gon, say ABC and CDE. Join AC and CE. Then2. Because the arcs AC and CE are equal, being sums of equal arcs, Chord AC chord CE. 3. Hence in the triangles ABC and CDE we have (§ 208) Therefore these triangles are identically equal, and Angle ABC angle CDE. = 4. In the same way it may be shown that any other two angles of the inscribed polygon we choose to take are equal. 5. Comparing (1) and (4), the polygon is shown to be regular (§ 152). Q.E.D. Scholium. The equality of the angles of the polygon may be proved with yet greater elegance by showing that they are all inscribed in equal segments. THEOREM II. 448. If a circumscribed polygon touch a circle at equidistant points around it, it is regular. Hypothesis. A circumscribed polygon whose sides touch the circle at the equidistant points ABCDE. Conclusion. This polygon has all its sides and angles equal. Proof. Let O be the centre of the circle. Join OA, OB, etc. Then 1. Because the intercepted arcs AB, BC, etc., are equal, we shall have D B A E Turn the figure around on the point O until the radius OA coincides with the trace OB. Then- 2. Because of the equality of the angles AOB, BOC, etc., OB will fall upon the trace OC; OC= trace OD, etc. 3. Because the radii are equal, the point A will fall on B, B on C, etc. 4. Because each radius is perpendicular to the tangent at its extremity, each side will fall upon the trace of the side next following. 5. Therefore each point of intersection will fall on the trace of the point next following. 6. Therefore each side and angle is equal to the side and angle next following, and the polygon has all its sides and angles equal. Q.E.D. Remark. Another demonstration may be found by drawing lines from 0 to the angles of the polygon, and proving the equality of all the triangles thus formed. |