very hard to draw and read.. but it is easy when you get a idea.

no ideas?

here's my drawing

I think Similarity and Menelaus will kill this problem $KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$ Then Menelaus, done.

Hypernova wrote: I think Similarity and Menelaus will kill this problem $KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$ Then Menelaus, done. nope How can you prove $KG$ is parallel to $QB$? One more, if you proved $KG//QB$, just done

khyeon wrote: Hypernova wrote: I think Similarity and Menelaus will kill this problem $KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$ Then Menelaus, done. nope How can you prove $KG$ is parallel to $QB$? read again

Skravin wrote: khyeon wrote: Hypernova wrote: I think Similarity and Menelaus will kill this problem $KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$ Then Menelaus, done. nope How can you prove $KG$ is parallel to $QB$? read again nothing changed

Hypernova wrote: I think Similarity and Menelaus will kill this problem $KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$ Then Menelaus, done. I assume that $RQ$ means $PQ$

Skravin wrote: Hypernova wrote: I think Similarity and Menelaus will kill this problem $KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$ Then Menelaus, done. I assume that $RQ$ means $PQ$ Same. Let $QP\cap BC=R$.

Hypernova wrote: Skravin wrote: Hypernova wrote: I think Similarity and Menelaus will kill this problem $KG // QB$ since $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ$ Then Menelaus, done. I assume that $RQ$ means $PQ$ Same. Let $QP\cap BC=R$. Ah got it

What I mean is $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ \to KG // QB$ I can't understand..

khyeon wrote: What I mean is $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ \to KG // QB$ I can't understand.. Full solution Let $QP\cap BC=R$. Since $CE=CD, RE=RP$ (Tangent), $\angle ECD=\angle ERP$. So $CK//RQ$. And $E, O_1, O_2$ collinear, so $\triangle EO_{1}D \sim EO_{2}P (AA)$. Hence $\frac{EK}{EQ}=\frac{ED}{EP}=\frac{O_1D}{O_2P}=\frac{EO_1}{EO_2}=\frac{EG}{EB}$, so $KG//QB$. Then use Menelaus in $\triangle EFN$, secant $LO_2M$. It gives $\frac{LF}{EL}*\frac{MN}{FM}*\frac{O_2E}{NO_2}=\frac{r_2}{r_1}*\frac{1}{2}*\frac{r_1}{r_2}=1$. So $r_2=2r_1$. Then $\frac{BG}{EG}=\frac{EO_2}{EO_1}=\frac{r_2}{r_1}=2$, Q.E.D.

khyeon wrote: What I mean is $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ \to KG // QB$ I can't understand.. It was very simple except the drawing (I didn't take the Junior exam)

Hypernova wrote: khyeon wrote: What I mean is $\triangle EO_{1}D \sim EO_{2}P (AA)$ and $CK // RQ \to KG // QB$ I can't understand.. It was very simple except the drawing (I didn't take the Junior exam) Thank you for your kindness... At my test, I proved $L$ is on the $O_1$

One of the best geometry questions I have ever seen. It is easy to assume that $ \triangle KGC \sim \triangle QBR $ The proof of this is the following :

So $ \bar{GL} \parallel \bar{BF}. $ Because of Menelaus, $ \frac{\bar{EL}}{\bar{LF}} \times \frac{\bar{FM}}{\bar{NM}} \times \frac{\bar{N{O}_{2}}}{\bar{E{O}_{2}}} =1$ So, $ \frac{\bar{EL}}{\bar{FL}}=\frac{2}{1} . $ So, $ \bar{BG}= 2 \times \bar{EG} $