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CTK Exchange
jmolokach
Member since Aug-17-10
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Aug-23-10, 03:01 PM (EST) |
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"Comment about proof #75."
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I recently created a diagram that contains the same diagram from proof 75 but extended the height to create a larger isosceles triangle, whose sides are 2d and base 2b. I will post a picture here when I get a chance, but one consequence of the diagram I noticed is that the 3 medians of an isosceles triangle divide the triangle into 6 triangles, each with equal area. I am sure this was already known...but new knowledge to me. The diagram also shows how to trisect the area of any right triangle. I know that a single median of any triangle divided the triangle into 2 smaller triangles equal in area, but do the three medians of ANY triangle create 6 smaller triangles of equal area? Also, I am wondering if anyone has made a connection between the PT and the centroid of a triangle. Success is getting up one more
time than you fall down... |
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alexb
Charter Member
2570 posts |
Aug-23-10, 03:08 PM (EST) |
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1. "RE: Comment about proof #75."
In response to message #0
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>I know that a single median of any triangle divided the triangle into 2 smaller triangles equal in area, but do the three medians of ANY triangle create 6 smaller triangles of equal area? Yes, in any triagle the three medians form six triangles of equal areas. This is ineed one of the characterizations of the centroid as the center of gravity. If G is the centroid of ΔABC and AD a median from A, then AD:GD = 1:3 so that Area(GBC) = Area(ABC)/3. Also, Area(ADG) = Area(BDG) = Area(GBC)/2 = Area(ABC)/6. |
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jmolokach
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Aug-24-10, 03:03 PM (EST) |
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4. "RE: Comment about proof #75."
In response to message #2
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I must admit to being in the wrong on this. The non-right triangles, although having equal areas do not lead to the Pythagorean Identity, unless someone sees something I don't, I will file this under "Ways not to build a light-bulb" category... |
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