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# Heard of cross-ratio?

In $$\Delta ABC$$, $$D$$, $$E$$, $$F$$ are such that $$B - D - C$$, $$C - E - A$$, $$A - F - B$$. $$\overline{AD}$$, $$\overline{BE}$$, $$\overline{CF}$$ are concurrent at $$P$$ which is in the interior of $$\Delta ABC$$. Ray $$FE$$ intersects ray $$BC$$ at $$N$$. $$\overline{AD}$$ intersects $$\overline{FE}$$ at $$M$$. Given $$FM = 3$$, $$ME = 2$$ find $$NE$$.

Can anyone give me a trigonometric solution for this?

This is part of the set Trigonometry.

Note by Omkar Kulkarni
3 years ago

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- 3 years ago

Can you tell me how you got the solution?

- 2 years, 11 months ago