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 $\square$

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 $\measuredangle DEC=\measuredangle BAG$