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In acute triangle $ABC$, $AB>AC$. Let $I$ the incenter, $\Omega$ the circumcircle of triangle $ABC$, and $D$ the foot of perpendicular from $A$ to $BC$. $AI$ intersects $\Omega$ at point $M(\neq A)$, and the line which passes $M$ and perpendicular to $AM$ intersects $AD$ at point $E$. Now let $F$ the foot of perpendicular from $I$ to $AD$. Prove that $ID\cdot AM=IE\cdot AF$

What I thought: Notice that $\triangle IDE$ and $\triangle AFM$ are similar after some angle chasing.

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  • $\begingroup$ If $IDE$ and $AFM$ are similar, then $ID/IE=AF/AM$, which is the relation you want to prove. $\endgroup$ Nov 16 '19 at 22:20
  • $\begingroup$ What is the meaning of "Mansion" in your title ? $\endgroup$
    – Jean Marie
    Nov 17 '19 at 0:45
  • $\begingroup$ This is question 2 from the 2019 Korean National Olympiad. $\endgroup$
    – user574848
    Nov 17 '19 at 10:27
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COMMENT:

You correctly found that $\triangle IDE≈\triangle AFM$.We rewrite the required relation as:

$\frac{AF}{AM}=\frac{ID}{IE}$

That is triangles IAE and DIE must be similar. These two triangles have common angle $\angle IED$, therefore we must have:

$\angle IDE=\angle EIA$

This is not possible because:

$\angle IDE=\angle BDE (=90^o)+\angle BDI$

$\angle EIA=\angle EIF (≠90^o)+\angle FIA(=BDI)$

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  • $\begingroup$ Great observation! How can I contact you? $\endgroup$
    – LockyPauk
    Nov 23 '19 at 15:34
  • $\begingroup$ through this site or this address (sirous_f20@yahoo.com) $\endgroup$
    – sirous
    Nov 23 '19 at 15:49

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