69. Definition. If a secant of a triangle pass through a vertex and the middle of the sides opposite, the segment between these points is called a median. There will be three medians in every triangle. Problems. - 1. Show that two medians of a triangle trisect each other; i.e. separate each other into two segments, one of which is one-third the whole median. The analysis of the problem suggests the following: Draw PQ, joining the middle points of AI and BI. It will be parallel to BA BA and equal to B FIG. 108. E Therefore PQ and DE are parallel and equal, and PQED is a parallelogram (Ex. 2, § 38). = Its diagonals bisect each other, or QI ID. But QI AQ by construction; therefore AQ = QI = ID. For the same reasons BP = PI = IE. Q. E. D. 2. Show that the third median would also pass through I. 3. Having given the three medians of a triangle, to construct it. APH will be a triangle, having the three medians for its sides. K will be the middle point of AP. Construction. - Form a triangle with the three medians as sides. Draw a median of this triangle through any vertex. Pro = duce it one-third of its length. Through either of the other vertices, as A, draw AQ and AR, and lay off QB AQ and RC AR. Join BC. ABC will be the required triangle. 4. If through any point three lines be drawn intersecting parallels, the segments will be proportional. 5. Establish the converse of Problem 4. E B P FIG. 111. 6. How does the bisector of an interior angle of a triangle divide the opposite side? If AD bisects the interior angle at A, we will have two triangles, BAD and CAD having an angle in each equal, and other two angles supplementary. Hence the opposite side is divided into segments proportional to the adjacent sides. 7. How does the bisector of an exterior angle of a triangle divide the opposite side? If AK bisects the exterior angle at A, we have two triangles, BAK and CAK having the ZK in common and the CAK supplementary to the By Ex. 2, § 65, BAK. BK AB = CK AC Hence the two segments formed by the point of intersection and the other vertices will be proportional to the sides having their vertex at the vertex of the bisected angle. BD BK 8. Show that = ᎠᏟ CK GENERAL EXERCISES. 1. On a given segment as one side construct a parallelogram similar to a given parallelogram. 2. Show that similar polygons may be separated by auxiliary lines into similar triangles. 3. Show that circles are similar figures. 4. Show that the corresponding altitudes of similar triangles will be proportional to any set of corresponding sides. FIG. 113. 5. Show that the radii of circles inscribed in similar triangles are proportional to the corresponding sides. 6. Show that the same relation exists between the diameters of circumscribed circles. 7. Show that the same relation exists between the radii of the corresponding escribed circles. 8. Show that the corresponding altitudes are to each other as the corresponding medians. 9. Show that the corresponding angle bisectors are to each other as the perimeters of the triangles. NOTE. There are three principles in the elements of geometry that are more prominent than any others. We have now established the first of these, as follows: Corresponding lines of similar figures are to each other as ANY OTHER corresponding lines. The second of these great principles, which will be established in the next chapter, is: Similar areas are to each other as the squares of any corresponding lines. The third, which will be established in Chapter XIII., is: Similar volumes are to each other as the cubes of any corresponding lines. CHAPTER VI. 70. THEOREM. The square constructed on the sum of two segments of a line equals the sum of the squares on the two segments PLUS twice the rectangle of the two segments. K G ab b2 H. Place the two segments so that MN shall be their sum. D E On MN as one side, construct the square MH. At Perect the 1 PG. On MK lay off MD = a. Draw DE to MN. show that: MDIP = a2, IGHE = b2, DKGI = ab, IENP = ab. Adding, we have (a+b)2= a2 + b2+2 ab; a relation that we are already familiar with in algebra. Q. E. D. 71. THEOREM. The square constructed on the difference of two segments of a line equals the sum of the squares on the two segments MINUS twice the rectangle of the two segments. Place the two segments so that MN shall be their difference, MP representing one segment a, and PN representing the other segment b. |