306.以△ABC的BC、CA、AB为底边分别在形外作三个相似的等腰三角形,使其底角为30°,顶点是O_1、O_2、O_3,求证:△O_1O_2O_3是正三角形。证:如图1,作C关于O_1O_2的对称点P,连PO_1、PO_2、PO_3、PA、PB、PC。由O_1B=O_1C=O_1P知O_1为△BPC的外心,∠PBC=1/2∠PO_1C,∠PCB=1/2∠PO_1B,故∠PBC+∠PCB=1/2(∠PO_1C+∠PO_1B)=1/2∠BO_1C=60°,∠BPC=120°。同理,∠CPA=120°。于是∠APB=120°。延长BO_3到Q,使O_3Q=O_3B,连QA。∵ 306. Using BC, CA, and AB of △ABC as the bottom edge, three similar isosceles triangles are formed outside the shape, respectively, so that the bottom angle is 30°, and the top points are O_1, O_2, and O_3. Prove that: ΔO_1O_2O_3 is an equilateral triangle. Card: As shown in Figure 1, C is the symmetry point P of O_1O_2, and even PO_1, PO_2, PO_3, PA, PB, and PC. O_1B=O_1C=O_1P knows O_1 is the heart of △BPC, ∠PBC=1/2∠PO_1C, ∠PCB=1/2∠PO_1B, so ∠PBC+∠PCB=1/2(∠PO_1C+∠PO_1B)=1 /2∠BO_1C=60°, ∠BPC=120°. Similarly, ∠CPA=120°. So ∠APB=120°. Extend BO_3 to Q so that O_3Q = O_3B, even QA. ∵