Theorem 33. The perpendiculars from the vertices of a triangle to the opposite sides are concurrent. NOTE. Br Z 3... B'CAB CA', and C is the mid-point of B'A'. Th. 24; ax. 1 4. Similarly, A and B are mid-points of B'C', C'A'. Th. 31 6. And they are also the perpendiculars to a, b, c. Th. 17, cor. 1 The theorem is due to Archimedes. The above, one of the simplest of many proofs suggested, is due to Gauss. DEFINITION. To trisect a magnitude is to cut it into three equal parts. EXERCISES. 214. Show that the perpendiculars to the sides of a triangle, not through their mid-points, (1) are not concurrent, (2) form a triangle equiangular to the given triangle, in general. 215. In th. 32 the Principle of Continuity suggests moving C upwards, or down to and through the side c. What becomes of P1, P2, P3, P4 if C moves down to rest on c? 216. Similarly, how are the points P affected if C moves through c, and stops below the line? 217. Similarly, if a becomes parallel to b? 218. If the opposite sides of a quadrilateral are perpendicular to each other, then two diagonals are perpendicular to each other. Such a quadrilateral is ABOC in the figure of th. 33. (Carnot's theorem.) Theorem 34. The medians of a triangle are concurrent in a trisection point of each. Given the ABC and the medians BY, AX, intersecting at O. To prove (1) that the median from C passes through O, (2) that OX = & AX, OYBY, etc. Proof. 1. Suppose CO drawn, and produced indefinitely, cutting AB at Z. 2. Suppose AP II OB; CO must cut Post. 5, cor. 3. Then. CYYA, ... CO = OP. 4. And CO = OP, and CX = XB, ... OX II PB. X B Th. 27, cor. 2 Th. 27, cor. 3 Th. 24, cor. 2 5... APBO is a, and AZ = ZB, and OZ = ZP. 6. .. CZ is a median, and it passes through O. = 1 7. And. OZ OP, it = CO, or CZ. Similarly for OY and OX.. DEFINITIONS. 1. The point of concurrence of the perpendicular bisectors of the sides of a triangle is called the circumcenter of the triangle. (Th. 31.) The reason will appear later when it is shown that this point is the center of the circum-scribed circle. (See Table of Etymologies.) 2. The point of concurrence of the bisectors of the interior angles of a triangle is called the in-center of the triangle; the points of concurrence of the bisectors of two exterior angles and one interior, are called the ex-centers of the triangle. (Th. 32.) It will presently be proved that the in-center is the center of a circle, in-side the triangle, just touching the sides; and that the ex-centers are centers of circles, out-side the triangle, just touching the three lines of which the sides of the triangle are segments. Hence the names in-center and ex-center. 3. The point of concurrence of the three perpendiculars from the vertices to the opposite sides is called the orthocenter of the triangle. (Th. 33.) 4. The point of concurrence of the three medians of a triangle is called the centroid of that triangle. (Th. 34.) It is shown in Statics that this point is also the center of mass, or center of gravity of the plane surface of the triangle. It is, therefore, sometimes called by those names. EXERCISES. 219. Draw a figure in which the circumcenter appears to fall on the side c; outside the triangle. What kind of a triangle does it appear to be? 220. May the in-center ever lie outside the triangle, as the circumcenter may? Proof. 221. If a triangle is acute-angled, prove that both the circumcenter and the orthocenter lie within the triangle. 222. Discuss the case of an obtuse-angled triangle, giving proof. (See ex. 221.) 223. Also of a right-angled triangle. (Transition stage between ex. 221 and ex. 222.) 224. In th. 34, if X, Y, Z be joined, the XYZ will be equiangular with the ABC. Prove this, and also investigate the relation of the sides of A XYZ to those of ▲ ABC, as to length. 225. Is there any kind of a triangle in which the in-center, circumcenter, orthocenter, and centroid coincide? If so, what is it? Prove it. 226. If from the vertex of a triangle where two unequal sides meet, there are drawn the bisector of the angle, the median, and the perpendicular to the opposite side, the first lies between the last two. 227. Suppose, in ex. 226, the unequal sides become equal. 228. The sum of the three medians of a triangle is greater than threefourths of its perimeter. 229. If two sides of a triangle are unequal, the median from their intersection makes the greater angle with the shorter of those sides. (In the figure of th. 34, produce CZ to M, making ZM : = CZ; join M and B.) 230. To trisect a given line. 231. In the figure of th. 34, connect X, Y, Z, and prove that O is also the centroid of A XYZ. 232. In ex. 231, prove that if the mid-points of the sides of ▲ XYZ are joined, O is also the centroid of that triangle; and so on. BOOK II. EQUALITY OF POLYGONS. Section 1. Theorems. DEFINITIONS. Two polygons are said to be adjacent if they have a segment of their perimeters in common. Suppressing the common segment of the perimeters of two adjacent polygons, a polygon results which is called the sum of the two polygons. Two surfaces which may be divided into the same number of parts respectively congruent, are said to be equal. A B This property is often designated by the expressions equivalent, equal in area, of equal content, etc.; but the use of the word congruent, for identically equal, renders the word equal sufficient, and this is the common meaning. (See definition of congruence, p. 20.) The definition is more broadly treated in Book V. The altitude of a trapezoid is the perpendicular distance between the base lines. Hence a trapezoid can have but one distinct altitude, a, unless it becomes a parallelogram, in which case it can have two, viz. a and a' in the annexed figure. The altitude of a triangle with reference to a given side as the base, is the distance from the opposite vertex to the line of that base. Hence a triangle can have three distinct altitudes, viz. a1, a2, as, in the figure. a ba a2 b b a b, a, a3 Theorem 1. Parallelograms on the same or equal bases and between the same parallels are equal. Given ABCD, ABC'D', on the same base AB, and be tween the same parallels P, P'. 5... ABCD= ABC'D', by taking 3 from 4. Ax. 3 6. Similarly for Figs. 2 and 3. NOTES. 1. In Fig. 2, CD' has become zero; in Fig. 3 it has passed through zero and become negative. Hence, considering the sense of these lines, i.e. their proper signs, the above proof holds for all figures. The meaning of "between the same parallels" is apparent from the figure. 2. From the definition of the sum of two polygons, p. 77, are readily Iderived the ideas of the difference of two and of the sum of several polygons. COROLLARIES. 1. A parallelogram equals a rectangle of the same base and the same altitude. For the rectangle is a special kind of parallelogram. 2. Parallelograms having equal bases and equal altitudes are equal. (Why ?) 3. Of two parallelograms having equal altitudes, that is the greater which has the greater base; and of two having equal bases, that is the greater which has the greater altitude. Why? 4. Equal parallelograms on the same or equal bases have equal altitudes. Law of Converse, after cors. 2 and 3. Give it in full. |