Let $ABC$ be a triangle with $\angle ABC=90^{\circ}$. We have: $1) BH \perp AC; $ $2)AD \text{ the bisector of } \angle{A} \text{ and } AD\cap BH=\{Q\},D\in BC;$
$3) CE \text{ the bisector of } \angle C \text{ and } CE \cap BH =\{P\},E \in AB; $ $4) CE \cap AD ={I};$ $5) NE=NP;$ $6) QM=MD;$
Prove that: $NM \parallel AC .$
This problem was proposed this year to National Olympiad from Romanian.
The solution can be check here: http://onm2012.isjcta.ro/doc/9_barem.pdf .
What I cannot understand is the the following relation:
$ \frac{QA}{QD}=\frac{c^2}{a^2}\cdot \frac{b+c}{c}.$
Thanks :)