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In triangle ABC, AB=10, CA=12. The bisector of ∠𝐁 intersects CA at E, and the bisector of ∠𝐂 intersects AB at D. AM and AN are the perpendiculars to CD and BE respectively. If MN=4, then find BC.

It isn't my homework, just a math problem that I randomly wanted to try out since I have no clue where to start with this.

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    $\begingroup$ I think the conditions are not enough. $MN=\frac{AB+AC-BC}{2}$ $\endgroup$ – Oolong milk tea May 11 at 10:25
  • $\begingroup$ @Oolongmilktea Could u plz tell me how u came to this? $\endgroup$ – Nabiha21 May 11 at 10:32
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    $\begingroup$ Let $U,V$ denote the intersection of $AM,BC$ and $AN,BC$. $MN=\frac{UV}{2}=\frac{BV+CU-BC}{2}=\frac{AB+AC-BC}{2}$. $\endgroup$ – Oolong milk tea May 11 at 10:38
  • $\begingroup$ why is MN= UV/2? Why is the denominator 2 tho? $\endgroup$ – Nabiha21 May 11 at 11:09
  • $\begingroup$ Notice $AM=MU AN=NV$. $\endgroup$ – Oolong milk tea May 11 at 11:12

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