Đáp án:

Giải thích các bước giải:

 a, $\triangle$ ABC có $\widehat{BAC}$ + $\widehat{ABC}$ + $\widehat{ACB}$ = $180^o$

=> $80^o$ + $\widehat{ABC}$ + $\widehat{ACB}$ = $180^o$

=> $\widehat{ABC}$ + $\widehat{ACB}$ = $100^o$

=> $\widehat{IBC}$ + $\widehat{ABI}$ + $\widehat{ICB}$ + $\widehat{ICA}$ = $100^o$

mà $\widehat{IBC}$ = $\widehat{ABI}$; $\widehat{ICA}$ = $\widehat{ICB}$ (GT)

=> 2($\widehat{IBC}$ + $\widehat{ICB}$) = $100^o$

=> $\widehat{IBC}$ + $\widehat{ICB}$ = $50^o$

- $\triangle$ IBC có $\widehat{IBC}$ + $\widehat{ICB}$ + $\widehat{BIC}$ = $180^o$

=> $50^o$ + $\widehat{BIC}$ = $180^o$

=> $\widehat{BIC}$ = $130^o$

b,

- Vì p/g $\widehat{ABC}$ và p/g $\widehat{ACB}$ cắt tại I (GT) 

=> IC = IB (t/c đg p/g trong tam giác) => $\triangle$ IBC cân tại I (đ/nghĩa)

=> $\widehat{IBC}$ = $\widehat{ICB}$ (t/c) 

mà $\widehat{IBC}$ = $\widehat{ABI}$; $\widehat{ICA}$ = $\widehat{ICB}$

=> $\widehat{IBC}$ = $\widehat{ABI}$ = $\widehat{ICA}$ = $\widehat{ICB}$

- Mặt khác: $\widehat{IBC}$ + $\widehat{ICB}$ = $50^o$ (cmt)

=> $\widehat{IBC}$ = $\widehat{ABI}$ = $\widehat{ICA}$ = $\widehat{ICB}$ = $25^o$

- $\triangle$ MCB có $\widehat{IBC}$ + $\widehat{MCB}$ + $\widehat{CMB}$ = $180^o$

=> $\widehat{IBC}$ + $\widehat{ICA}$ + $\widehat{ICB}$ + $\widehat{CMB}$ = $180^o$

=> $25^o$ . 3 + $\widehat{CMB}$ = $180^o$

=> $75^o$ + $\widehat{CMB}$ = $180^o$

=> $\widehat{CMB}$ = $125^o$ 

Ta có: $\widehat{CMB}$ = $125^o$ ; $\widehat{BIC}$ = $130^o$ ; $\widehat{BAC}$ = $80^o$

=> $\widehat{BIC}$ > $\widehat{CMB}$ > $\widehat{BAC}$

#Chii

#Team: Extensive Knowledge