2010年5月号问题解答(解答由问题提供人给出)1851在△ABC中,以B,C为焦点的椭圆分别交边AB,AC于点E,D.证明:△ABD与△ACE有相同的内心.(山东宁阳第一中学刘才华271400)证明设∠BAC=A,∠ABC=B,∠ACB=C,∠DBC=α,∠ECB=β,△ABC,△ABD,△ACE的内心分别为I,I_1,I_2,因它们都在∠A平分线AI上,则点I,I_1,I_2共线,I_1,I_2在点I的同侧.在△BII_1中,∠BI_1I=∠ABI_1+∠BAI_1=(A+B-α)/2,由正弦定理得 May 2010 Problem-solving (answer provided by the person providing the problem) 1851 In △ ABC, the ellipses with B and C as the focal points respectively intersect AB and AC at points E and D. It is proved that △ ABD and △ ACE have The same inside. (Shandong Ningyang First Middle School Liu Cahua 271400) Prove that ∠ BAC = A, ∠ ABC = B, ∠ ACB = C, ∠ DBC = α, ∠ ECB = β, △ ABC, △ ABD, Are I, I_1 and I_2, respectively, because they are all on the ∂A bisector AI, the points I, I_1 and I_2 are collinear, and I_1 and I_2 are on the same side of point I. In △ BII_1, ∂BI_1I = ∠ ABI_1 + ∠BAI_1 = (A + B-α) / 2, obtained by the sine theory