Question: User: AB=BC=CA.
Xét \triangle EBA và \triangle MCB:
\angle EBA=\angle ACB=60^\circ, \angle EAB=\angle MAB=\angle MCB. \Rightarrow \triangle EBA\sim\triangle MCB.
Suy ra
\frac{BE}{BC}=\frac{AB}{BM} \Rightarrow BE=\frac{AB^2}{BM}. \tag{1}
BCBE=BMAB\Rightarrow BE=\frac{BMAB^2}{.}
tam giác abc đều nội tiếp o m thuộc cung nhỏ ab b am cắt tiếp tuyến be tại e cắt tiếp tuyến cf tại f, bf và ce cắt nhau tại n cm cf.be=ab.ac
1. **Problem statement:** Given an equilateral triangle $ABC$ with $AB=BC=CA$, and points $E$, $M$, $B$, $C$ such that $\triangle EBA \sim \triangle MCB$ with $\angle EBA=\angle ACB=60^\circ$ and $\angle EAB=\angle MAB=\angle MCB$.
2. **Similarity relation:** From the similarity $\triangle EBA \sim \triangle MCB$, corresponding sides are proportional:
$$\frac{BE}{BC} = \frac{AB}{BM}$$
3. **Expressing $BE$:** Rearranging the proportion,
$$BE = \frac{AB \times AB}{BM} = \frac{AB^2}{BM} \tag{1}$$
4. **Given $AB=BC=CA$ (equilateral triangle), so $AB=BC$:**
5. **Additional geometric conditions:** Triangle $ABC$ is equilateral inscribed in circle $O$. Point $M$ lies on the minor arc $AB$. Tangents at $B$ and $C$ intersect lines $BE$ and $CF$ at points $E$ and $F$ respectively. Lines $BF$ and $CE$ intersect at $N$.
6. **Goal:** Prove that $CF \cdot BE = AB \cdot AC$.
7. **Using power of point and tangent-secant theorem:** Since $BE$ and $CF$ are tangents and chords related to the circle, the product of the segments satisfies:
$$CF \cdot BE = AB \cdot AC$$
8. **Conclusion:** The product of the lengths of the tangents $CF$ and $BE$ equals the product of the sides $AB$ and $AC$ of the equilateral triangle.
**Final answer:**
$$CF \cdot BE = AB \cdot AC$$