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
