A geometry problem proposed at national olympiad. 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 :) 

See the figure below:

Let $K$ be a point such that $K \in BQ$ and $\angle QKD$ is a right angle.
Using similarity and angle bisector theorem we get:
$$\frac{AH}{HC}=\frac{c^2}{a^2} \quad(1)$$
and
$$\frac{DB}{DC}=\frac{c}{b}. \quad(2)$$
From equation $(2)$ we conclude that
$$\frac{DB}{BC}=\frac{c}{b+c}. \quad(3)$$
Note that $\triangle BDK \sim \triangle BCH$, therefore
$$\frac{DB}{BC}=\frac{KD}{HC}. \quad(4) $$
From $(3)$ and $(4)$ we get:
$$\frac{KD}{HC}=\frac{c}{b+c}. \quad(5)$$
Dividing $(1)$ by $(5)$ we get:
$$\frac{AH}{KD}=\frac{c^2}{a^2} \cdot \frac{b+c}{c}. \quad(6)$$
But as $\triangle AHQ \sim \triangle DKQ$, we can conclude that
$$\frac{QA}{QD}=\frac{AH}{KD}= \frac{c^2}{a^2} \cdot \frac{b+c}{c}$$
