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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.

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

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

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