The thereom of mean similiarty.

by pythag011, Sep 17, 2008, 12:46 AM

Let's say we have two DIRECTLY(whcih means both clockwise or counterclockwise) similiar triangles. ABC and DEF. Let X be the midpoint of AD, Y be teh midpoint of BE and Z be the midpoint of CF.

Then, XYZ is similiar to ABC and DEF.

This can be very useful.

BTW, to explain DIRECTLY similiar and not directly similiar, these triangles are directly similiar

[geogebra]743d4b453524ba09db74b2b69ec2a5521edab340[/geogebra]

The ones on the left are directly similar. The ones on the right are oppositely similar.

More generaly: Let there be two directly similiar polygons. Then teh polygon formed by their midpoints of corresponding vertices are directly similiar to th epolygons.

Yeah, I know I suck at exlaining things. *sigh* :wallbash:

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4 Comments

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Directly similar: If, on a plane, they have the same chirality, they are directly similar.
That is to say if the only tools you need to obtain the two triangles are homothety and rotation
Oppositely similar: If you need to flip the diagram in order to obtain the similarity, they are oppositely similar.
Like you hands... somewhat. Rotate your hands, no matter how hard you try you WON'T MATCH THEM UP.

by james4l, Sep 17, 2008, 1:08 AM

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dude we can like prove this by taking the center of spiral similarity between these two triangles (polygons) and then like use spiral similarity to get the desired direct similarity

by The QuattoMaster 6000, Sep 17, 2008, 3:20 AM

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hello

vectors

by not_trig, Sep 19, 2008, 1:15 AM

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hello spiral similarity

by The QuattoMaster 6000, Sep 19, 2008, 10:51 PM

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