\(a)\) Ta có: \(OB // O’C\;\;(gt)\)
Suy ra: \(\widehat {AOB} + \widehat {AO'C} = 180^\circ \) (hai góc trong cùng phía)
\(OA = OB ( = R)\)
\(⇒\) Tam giác \(AOB\) cân tại \(O.\)
Suy ra: \(\widehat {BAO} = \displaystyle {{180^\circ - \widehat {AOB}} \over 2}\)
\(O'A = O'C ( = R)\)
\(⇒\) Tam giác \(AO'C\) cân tại \(O'\)
Suy ra: \(\widehat {CAO'} = \displaystyle{{180^\circ - \widehat {AO'C}} \over 2}\)
Ta có: \(\displaystyle\widehat {BAO} + \widehat {CAO'}\)\(\displaystyle = {{180^\circ - \widehat {AOB}} \over 2} + {{180^\circ - \widehat {AO'C}} \over 2}\)
\(\displaystyle = {{180^\circ + 180^\circ - (\widehat {AOB} + \widehat {AO'C})} \over 2}\)
\(\displaystyle = {{180^\circ + 180^\circ - 180^\circ } \over 2} = 90^\circ \)
Lại có: \(\widehat {BAO} + \widehat {BAC} + \widehat {CAO'} = 180^\circ \)
Suy ra: \(\widehat {BAC} = 180^\circ - (\widehat {BAO} + \widehat {CAO'})\)
\( = 180^\circ - 90^\circ = 90^\circ \)
\(b)\) Trong tam giác \(IBO,\) ta có: \(OB // O'C\)
Suy ra: \(\displaystyle{{IO'} \over {IO}} = {{O'C} \over {OB}}\) ( hệ quả định lí Ta-lét)
Suy ra: \(\displaystyle{{IO'} \over {IO}} = {1 \over 3} \Rightarrow {{IO - IO'} \over {IO}}\)
\(\displaystyle = {{3 - 1} \over 3} \Rightarrow {{OO'} \over {IO}} = {2 \over 3}\)
Mà \(OO’ = OA + O’A = 3 + 1 = 4 (cm)\)
Suy ra: \(\displaystyle{4 \over {IO}} = {2 \over 3} \)\(\displaystyle \Rightarrow IO = {{4.3} \over 2} = 6 (cm).\)
