## Subject II. (30 points)

In triangle $\triangle A B C$, the angle bisector $[A E, \mathrm{E} \in(B C)$ intersects the median $[B F], \mathrm{F} \in(A C)$, at point G.

a) Determine $a \in \mathrm{N}$, such that $\frac{B G}{G F} \cdot \frac{C E}{B E}=2^{a}$.

b) If triangles $\triangle A G F$ and $\triangle B G E$ are equivalent, then G is the centroid of $\triangle A B C$.

Prof. Elena Măgdaş, Horea Gymnasium School, Cluj-Napoca
## Subject II. (30 points)

Correct drawing
a) $\left[A E\right.$ bisector $\Rightarrow \frac{B E}{E C}=\frac{A B}{A C}$

$\left[A G\right.$ bisector $\Rightarrow \frac{B G}{G F}=\frac{A B}{A F} \Rightarrow \frac{B G}{G F}=\frac{A B}{\frac{A C}{2}} \Rightarrow \frac{B G}{G F}=\frac{2 A B}{A C} \quad ; \quad \frac{B G}{G F} \cdot \frac{C E}{B E}=\frac{2 A B}{A C} \cdot \frac{A C}{A B}=2 \Rightarrow a=1$.
b) $A_{\triangle B E G}=A_{\triangle B E A}-A_{\triangle B G A}$

$$
A_{\triangle A G F}=A_{\triangle B F A}-A_{\triangle B G A} \Rightarrow A_{B E A}=A_{\triangle A F B} \Rightarrow \frac{B A \cdot E M}{2}=\frac{B A \cdot F N}{2} \Rightarrow E M=F N
$$

$E M \perp A B, F N \perp A B \Rightarrow E M\|F N \Rightarrow E F\| A B, F$ midpoint of segment $\quad A C \Rightarrow E \quad$ midpoint of segment $B C \Rightarrow G$ centroid of $\triangle A B C$.