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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 :)

1 Answers 1

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See the figure below:

ABC triangle

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 $\frac{QA}{QD}=\frac{AH}{KD}= \frac{c^2}{a^2} \cdot \frac{b+c}{c}$

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    @JohnD. Thanks :)2012-12-31