B dagi inversiyadan keyin masala quyidagicha bo'ladi: BO1 - balandlik, BI1 - bissektrisa, markazi C1 bo'lgan doira. (!) (BA1D1) I1O1 ga tegadi. Burchaklarni aylantiramiz: burchak(A1BC1) = alfa, keyin burchak(A1C1I1) = 2burchak(A1BI1) = 2alfa, keyin burchak(C1A1I1) = 90 - alfa=> burchak(D1A1I1) = 90 + alfa, lekin burchak( A1I1D1) = burchak(A1BO1) = 90 - 2alfa, keyin A1D1I1 burchaklar yig'indisi bilan bizda burchak (A1D1I1) = 180 - (90 + alfa) - (90 - 2alfa) = alfa = burchak (A1BI1), shuningdek burchak (A1I1B) = burchak (BC1A1) /2 = (180 - 4alfa) : 2 = 90 - 2alfa = burchak(A1I1D1), keyin A1I1B va A1I1D1 uchburchaklar teng => BA1 = D1A1 => burchak(D1BA1) = burchak(BA1C1) /2 = alfa = burchak (A1D1I1), shuning uchun teginish.

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