Triangle Construction Given MediansDate: 12/12/2001 at 09:51:43 From: Chandra Murlaidhar Subject: Triangles - construction given medians only The lengths of the three medians are given. Construct a triangle. The median lengths are 5, 6, and 7. The steps given in the book are: Draw triangle ADE with sides DE = 5, AD = 6, AE = 7. Draw medians AF and DG of sides DE and AE of triangle ADE, intersecting each other at P. Extend AF to B so that PF = FB. Join BE and produce it to C so that EC = BE. Join AC. ABC is the required triangle.. The explanation as to how this is derived is not clear. Please explain the principle behind this construction. Thanks. Chandra Date: 12/12/2001 at 13:02:37 From: Doctor Peterson Subject: Re: Triangles - construction given medians only Hi, Chandra. You might see this better if you work backward; this is probably how one would derive the construction initially. Draw a triangle ABC (which we are to contruct), and its medians AE, CP, and BQ: Now we want to construct a triangle whose sides are congruent to the medians of ABC. The trick is to notice that if we slide the medians around (keeping each of them parallel), they will form a triangle. So we'll slide CP left to ED, and slide BQ up to DA, keeping median AE in place. Of course, this isn't a proper construction yet, and needs proof. We need to find point D such that ED = CP and DA = BQ. To do this, note that if point F is the midpoint of BP, then EF is parallel to CP and half its length. Then if we double FE to DE, ED = CP and is parallel to it. DP will be congruent and parallel to EC; and since P is the midpoint of AB, G (where DP intersects AE) is the midpoint of AE. We just have to show that DA = BQ. There are several ways I can see to show this; one is to see that triangles CEP and EBD are congruent, so that BD = EP are parallel; but EP is parallel to CA and half its length, so EP = QA, and we have yet another parallelogram BDAQ, from which we see that DA = BQ. As I said at the start, this is all a matter of parallel slides, which are represented in the proof by parallelograms. So we construct ADE, the triangle whose sides are equal to the medians of ABC, by taking P as the midpoint of AB and F as the midpoint of PB, then doubling EF to find D. P is then the centroid (intersection of the medians) of ADE. Now your construction just takes this in reverse. Given ADE, we find P, the intersection of medians AF and DG; then double PF to find B and double BE to find C. That should help you see the thinking involved here. If you need any more help with it, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/