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Pretty much stuck at where I should go. Any guidance?

http://i.stack.imgur.com/jbllI.jpg

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closed as off-topic by Jack, zz20s, Robert Z, JMP, Claude Leibovici Dec 12 '17 at 10:56

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Knowing the angle bisector theorem and denoting $BC = x$ you just need to note that $$\frac{6}{x}=\frac{4}{15-x}.$$ Solve for $x=BC$.

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  • $\begingroup$ +1 I didn't know whether to assume the OP knew the theorem. Spot on! $\endgroup$ – Namaste Jan 25 '13 at 3:57
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Use the law of sines in triangle ABD and also in triangle BDC; use what you know about $AB+BC$, what you know about angles ABD and DBC, and what you know about angles ADB and BDC.

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Hints:

  • Use the law of sines for each of triangles $\triangle ABD$ and $\triangle BDC$.

  • You know that $AB + BC + CA = 25$ ( = perimeter), and

  • $AC = 4 + 6 = 10$, so $AB + BC = 25 - 10 = 15$

  • You know that $\angle ABD \cong \angle CBD$, and the $\angle ADB + \angle BDC = 180^\circ$ (as they are are supplementary).

You have all the information you need to solve for $BC$

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  • $\begingroup$ Nice to see you too, @Babak. I'm assuming our time zones differ significantly! ~08:00 my time, now! $\endgroup$ – Namaste Feb 18 '13 at 13:53

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