Câu 1:
\(\widehat{BIC}=\frac{\text{sđc(BC)}-\text{sđc(AD)}}{2}\)
Câu 2:
\(\widehat{DIE}=\frac{\text{sđc(EmD)}+\text{sđc(CnF)}}{2}\)
Câu 3:
Ta có: \(\text{sđc(BC nhỏ)}=2\widehat{BDC}; \text{sđc(AD nhỏ)}=2\widehat{ABD}\)
\(\Rightarrow \widehat{BSC}=\frac{\text{sđc(BC nhỏ)}-\text{sđc(AD nhỏ)}}{2}=\frac{2\widehat{BDC}-2\widehat{ABD}}{2}=\widehat{BDC}-\widehat{ABD}=45^0-30^0=15^0\)
Bài 4:
Ta có:
\(\text{sđc(AD nhỏ)}=2\widehat{ACD}; \text{sđc(BC nhỏ)}=2\widehat{BAC}\)
\(\widehat{AMB}=180^0-\widehat{AMD}=180^0-\frac{\text{sđc(AD nhỏ)}+\text{sđc(BC nhỏ)}}{2}\)
\(=180^0-\frac{2\widehat{ACD}+2\widehat{BAC}}{2}=180^0-(\widehat{BAC}+\widehat{ACD})\)
\(=180^0-(30^0+30^0)=120^0\)