Source of the problem: 请问这个题能用GGB画出来吗？利用相似巧妙转化，兼谈喻平教授对数学教育研究的三点重要看法
When weakening one condition, the problem becomes:
In triangle ABC, ∠A=60°, and D is a point on BC such that BD:CD=1:2. Given AD=2, find the minimum value of BC.
Solution (using rotation transformation): Extend BE parallel to AC from point B, intersecting the extended line AD at point E. In triangle ABE, we have :
AE=AD+DE=2+1=3 and ∠ABE=120°.
Thus, the trajectory of point B is an arc of a circle, as shown in the figure below :
Since D is the trisection point of chord AE, we find OD=1. Extend OD to point I so that DI=2OD=2, making OI=3. Draw line IC, as shown in the figure below :
From triangles BDO and CDI being similar, we have
IC=2√3 (a constant value).
Therefore,
point C lies on circle I with center I and radius 2√3.
Let OI intersect circle O at point K. Clearly, when point B reaches K, BC is minimized, as illustrated in the figure below :
At this time,
BC=IC-IB(IK)=2√3-(3-√3)=3√3-3.
Thus,
BC≥3√3-3,
with equality achieved if and only if ∠ABC=45° (Because triangle AOI is a right triangle. )
