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Distances from a Point inside an Equilateral Triangle

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Date: 12/09/2001 at 17:33:58
From: Russ Bunce
Subject: Distances from a point inside an Equilateral Triangle

Dr. Math,

I am attempting to prove that the sum of the distances from a point
inside an equilateral triangle, measured parallel to the sides, is
equal to the length of the side of the triangle. I can prove this from
the "center" of the triangle, but I'm not sure how to prove this for
any point inside the triangle.

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```
Date: 12/10/2001 at 14:13:57
From: Doctor Floor
Subject: Re: Distances from a point inside an Equilateral Triangle

Hi, Russ,

We have to be very careful about how we measure these distances.

To be precise, in this figure you want to show that DI + DK + DL is
equal to the sidelength, or DH + DJ + DM. But, for instance, not
DI + DH + DL.

To prove this, we consider the figure, and note that IHD, LMD and KJD
are equilateral triangles.

Also we note that the altitudes of these three equilateral triangles,
DE, DF, and DG, are also the altitudes of triangles ADC, ADB, and BDC
(indicated by the red dotted segments). These three triangles together
are exactly ABC. Since the bases to the altitudes DE, DF, and DG in
ADC, ADB, and BDC are all equal to the sidelength of ABC, a, we get

a/2*AN = [ABC]
= DE*a/2 + DF*a/2 + DG*a/2
= a/2*(DE + DF + DG)

This yields that the sum of altitudes DE+DF+DG is equal to the
altitude of ABC.

By triangle similarity the sidelengths of equilateral triangles are
all a fixed number (exactly 2/3*sqrt(3)) times the lengths of the
corresponding altitudes. But that means that we can conclude that

a = HI + JK + LM
= DI + DK + DL
= DH + DJ + DM

as desired.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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