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Triangle Construction Given Medians

Date: 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.


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 

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 

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   
Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons

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