# Mansion point and incenter

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.

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

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:

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

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

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