## Sunday, January 17, 2010

### Pythagoras extended

I came across an interesting Almost Pythagoras relation at Pat'sBlog. It says that for any triangle ABC with median AM drawn from vertex A, we have the general relation:

AB² + AC² = BM² + AM² + MC² + AM².

This relation can be proven with the law of cosines.

I like the alternative proof which is derived from the fact that for any triangle ABC:

AB² + AC² = constant,

if point A is located on a circle concentric with the circle of diameter BC (see Figure 1).

A proof of this important (and practically unknown) triangle theorem is given by Nguyen Tan Tai at one of his pages.

Therefore, from Figure 1, we can draw the isosceles triangle A'BC of Figure 2 with AB² + AC² = A'B² + A'C² and median A'M = AM.

The relation A'B² + A'C² = (BM² + MA²) + (MC² + MA²) is then easily read from the Pythagorean relation on both right triangles A'MB and A'MC.

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