Nordic MC 2008 Q3

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#1 h Mar 3, 2013, 12:21 PM • 2 Y

Y by Adventure10 and 1 other user

Let $ABC$ be a triangle and $D,E$ be points on $BC,CA$ such that $AD,BE$ are angle bisectors of $\triangle ABC$ . Let $F,G$ be points on the circumcircle of $\triangle ABC$ such that $AF||DE$ and $FG||BC$ . Prove that $\frac{AG}{BG}= \frac{AB+AC}{AB+BC}$ .

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tukhtn

#2 h Mar 3, 2013, 2:06 PM • 2 Y

Y by Adventure10, Mango247

ACCCGS8 wrote:

Let $ABC$ be a triangle and $D,E$ be points on $BC,CA$ such that $AD,BE$ are angle bisectors of $\triangle ABC$ . Let $F,G$ be points on the circumcircle of $\triangle ABC$ such that $AF||DE$ and $FG||BC$ . Prove that $\frac{AG}{BG}= \frac{AB+AC}{AB+BC}$ .

We have $\dfrac{AB}{BC}=\dfrac{AE}{EC}$ .Hence, $\dfrac{AB+BC}{BC}=\dfrac{AE+EC}{EC}$ , or $AB+BC=\dfrac{AC.BC}{EC}$

Similarly, we can prove $AB+AC=\dfrac{BC.AC}{CD}$

So it implies to prove that: $\dfrac{AG}{BG}=\dfrac{EC}{CD}$

Accroding to the assumption, it's easy to show that $\measuredangle DEC=\measuredangle FAG$ .On the other hand, $\measuredangle ECD=\measuredangle AGB$ .
Thus $\Delta AGB$ is similar to $\Delta ECD$ , so $\dfrac{AG}{BG}=\dfrac{EC}{CD}$

The problem is completely solved $\square$

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#3 h Mar 6, 2015, 4:31 PM • 1 Y

Y by Adventure10

tukhtn wrote:

ACCCGS8 wrote:

Let $ABC$ be a triangle and $D,E$ be points on $BC,CA$ such that $AD,BE$ are angle bisectors of $\triangle ABC$ . Let $F,G$ be points on the circumcircle of $\triangle ABC$ such that $AF||DE$ and $FG||BC$ . Prove that $\frac{AG}{BG}= \frac{AB+AC}{AB+BC}$ .

We have $\dfrac{AB}{BC}=\dfrac{AE}{EC}$ .Hence, $\dfrac{AB+BC}{BC}=\dfrac{AE+EC}{EC}$ , or $AB+BC=\dfrac{AC.BC}{EC}$

Similarly, we can prove $AB+AC=\dfrac{BC.AC}{CD}$

So it implies to prove that: $\dfrac{AG}{BG}=\dfrac{EC}{CD}$

Accroding to the assumption, it's easy to show that $\measuredangle DEC=\measuredangle FAG$ .On the other hand, $\measuredangle ECD=\measuredangle AGB$ .
Thus $\Delta AGB$ is similar to $\Delta ECD$ , so $\dfrac{AG}{BG}=\dfrac{EC}{CD}$

The problem is completely solved $\square$

$\measuredangle DEC=\measuredangle FAG$ is wrong
the correct one $\measuredangle DEC=\measuredangle BAG$

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