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 :)