Distances from a Point inside an Equilateral TriangleDate: 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. Thanks for any help you can provide. Date: 12/10/2001 at 14:13:57 From: Doctor Floor Subject: Re: Distances from a point inside an Equilateral Triangle Hi, Russ, Thanks for your question. 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] = [ADC] + [ADB] + [BDC] = 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/ |
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