Median geometry help11/12/2023 ![]() In other words, D = M c, the midpoint of AB.Īs Dani points, his proof does not rely on the fact that the common point of the three medians divides each in the ratio 2:1, but rather obtains that ratio as a side effect. In any parallelogram, the diagonals bisect each other. With two pairs of parallel sides, the quadrilateral AHBG is a parallelogram. ![]() In ΔACH, M b is the midpoint of AC, while G is the midpoint of CH. In ΔBCH, M a is the midpoint of BC and also BH||GM a. Draw through B a line parallel to AM a and let it intersect CG at H. Extend CG beyond G and further beyond its intersection with AB at D. Let G be the point of intersection of the medians AM a and BM b. Smith Jewish Day School, Rockville, MD, which I came across in Mathematics Teacher, v 96, n 6, Sept. The fourth proof is due to Dani Rubinstein, Charles E. Since BC/M bM c = 2, we also have BG/M bG = 2 and CG/M cG = 2. All their sides are in the same proportion. The triangles BCG and M bM cG are similar. The proof ends as before.Ī third elementary proof was suggested by Scott Brodie. Therefore, GM b and GM C are twice as short as respectively GB and GC. This implies that the quadrilateral SRBC is a parallelogram, so that its diagonals CR and BS are halved by their common point G. Then M bM c is the mid-line in both triangles ABC and GRS, whence SR = BC and the lines are parallel. Similarly, CM c extends beyond M c to the length of GM c. Since for two medians the points thus defined coincide, the same, by symmetry, is true for the remaining median.Īnother elementary proof starts with extending BM b beyond M b to the length of GM b. Since diagonals of a parallelogram are halved by the point of intersection, we also have M bG = GS and M cG = GR which together with GS = SB and GR = RC show that, on both medians, G stands twice as far from one end (the vertex) than from the other end (the mid-point.) This condition uniquely defines a point on a median. Furthermore, the two lines are parallel (being both parallel to BC.) Thus, the quadrilateral M bM cSR is a parallelogram. Let also R be the mid-point of GC and S the mid-point of GB. Let G denote the point of intersection of BM b and CM c. the line joining the mid-points of two sides of a triangle, is parallel to the third side and equals its half (see, Euclid, Elements, VI.2 and VI.4.) Therefore, from ΔABC, M bM c = BC/2. Three medians of a triangle meet at a point - centroid of the triangle.īelow are several proofs of this remarkable fact.Ī mid-line, i.e. Using the standard notations, in ΔABC, there are three medians: AM a, BM b, CM c. In a triangle, a median is a line joining a vertex with the mid-point of the opposite side.
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