about dsargues' theorem
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Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
we know that 2 simlar triangles are perspective from a point,we get the point by joining the corresponding points,but consider 2 similar triangles $\triangle ABC,\triangle ADE$,which mans they have a common vertex(they are similar in the given orientation),so how do i get the line corresponding from $A$ without locating the point of perspectivity.
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Re: about dsargues' theorem
The point of perspectivity is defined by the lines joining $B,D$ and $C,E$ then , as the line $AA$ has no meaning.
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Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: about dsargues' theorem
there are special cases of Pascal's theorem where 2 coinciding points of a hexagon are considered as tangents.can this be also done here?
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Re: about dsargues' theorem
But here, you will draw tangents... to what?
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
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