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设A_1,B_1,C_1分别是△ABC中BC,CA,AB边上的任意点,则你△A_1B_1C_1为△ABC的内接三角形。本文中记△ABC的面积为S,AB=c,BC=a,CA=b,内切圆半径为r,三旁切圆半径为r_a,r_b,r_c;AC_1/C_1B=m,BA_1/A_1C=n,CB_1/B_1A=l,△AC_1B_1,△BA_1C_1,△CB_1A_1,△A_1B_1C_1的面积分别为S_1,S_2,S_3,S′。则有。定理、△ABC的面积S与其内接△A_1B_1C_1面积S′有如下关系式:S′=(1+mnl)/((1+m)(1+n)(1+l))S其中AC_1/C_1B=m,CB_1/B_1A=l,BA_1/A_1C=n。
Let A_1, B_1, and C_1 be any points on the BC, CA, and AB sides of △ABC, respectively, then you △A_1B_1C_1 is the intrinsic triangle of △ABC. In this paper, the area of △ABC is S, AB=c, BC=a, CA=b, the radius of inscribed circle is r, and the radius of three sides is r_a, r_b, r_c; AC_1/C_1B=m, BA_1/A_1C =n, CB_1/B_1A=l, △AC_1B_1, △BA_1C_1, △CB_1A_1, △A_1B_1C_1 are S_1, S_2, S_3, and S′, respectively. There are. Theorem, △ABC area S and its inward △A_1B_1C_1 area S′ has the following relationship: S′=(1+mnl)/((1+m)(1+n)(1+l))S where AC_1/ C_1B=m, CB_1/B_1A=l, BA_1/A_1C=n.