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.

#Geometry

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Neeraj Meena - 6 years, 4 months ago

Can you tell me how you got the solution?

Omkar Kulkarni - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$