Giải thích các bước giải:

a.Xét $$\Delta OAN,\Delta OBM$$ có:

$$ON=OM$$

Chung $$\hat O$$

$$OA=OB$$

$$\to\Delta OAN=\Delta OBM(c.g.c)$$

b.Ta có: $$AM=OA-OM=OB-ON=BN$$

c.Từ câu a $$\to\widehat{OBM}=\widehat{OAN},\widehat{OMB}=\widehat{ONA}$$

$$\to\widehat{IBN}=\widehat{IAM},\widehat{INB}=180^o-\widehat{ONA}=180^o-\widehat{BMO}=\widehat{IMA}$$

Xét $$\Delta IBN,\Delta IAM$$ có:

$$\widehat{IBN}=\widehat{IAM}$$

$$BN=AM$$

$$\widehat{IMA}=\widehat{INB}$$

$$\to\Delta IAM=\Delta IBN(g.c.g)$$

$$\to IA=IB, IM=IN$$

Xét $$\Delta OIM,\Delta OIN$$ có:
Chung $$OI$$

$$OM=ON$$

$$IM=IN$$

$$\to\Delta OIM=\Delta OIN(c.c.c)$$

$$\to\widehat{IOM}=\widehat{ION}$$

$$\to OI$$ là phân giác $$\widehat{xOy}$$

d.Từ câu a $$\to IA=IB$$

    Vì $$K$$ là trung điểm $$AB\to KA=KB$$

$$\to OA=OB, IA=IB, KA=KB$$

$$\to O, I, K\in$$ trung trực $$AB$$

$$\to O, I, K$$ thẳng hàng