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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 :)

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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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  • $\begingroup$ How did you generate the figure? TikZ? $\endgroup$
    – JohnD
    Dec 31 '12 at 2:06
  • $\begingroup$ @JohnD. No, John, I used C.a.R. (Compass and Ruler) and then changed the fonts in Serif Draw Plus. $\endgroup$ Dec 31 '12 at 2:19
  • $\begingroup$ Well it looks great. Very polished and professional looking. $\endgroup$
    – JohnD
    Dec 31 '12 at 3:26
  • $\begingroup$ @JohnD. Thanks :) $\endgroup$ Dec 31 '12 at 10:38

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