\(\widehat{ABC}+\widehat{ABD}=180^o\)
\(50^o+\widehat{ABD}=180^o\)
\(\widehat{ABD}=180^o-50^o=130^o\)
Vì \(\widehat{ACE}\) là 1 TH tượng tự của \(\widehat{ABD}\)
\(\Rightarrow\widehat{ABD}=\widehat{ACE}=130^o\)
Ta có: \(AB=BD=AC=EC\)
\(\Rightarrow\Delta ABD\) cân tại B và \(\Delta ACE\) cân tại C
\(\Rightarrow\widehat{BAD}=\widehat{BDA}=\widehat{CAE}=\widehat{CEA}=\frac{\widehat{ABD}\left(\widehat{ACE}\right)}{2}=\frac{180^o-130^o}{2}=25^o\)
***
Xét tam giác ABC có:
\(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^o\)
\(\widehat{BAC}+50^o+50^o=180^o\)
\(\widehat{BAC}=180^o-50^o-50^o=80^o\)
Ta có: \(\widehat{DAE}=\widehat{BAC}+\widehat{BAD}+\widehat{CAE}=80^o+25^o+25^o=130^o\)
Hoặc giairi từ ngay chỗ *** ấy là:
Xét tam giác \(ADE\) có:
\(\widehat{BDA}+\widehat{CEA}+\widehat{DAE}=180^o\)
\(25^o+25^o+\widehat{DAE}=180^o\)
\(\widehat{DAE}=180^o-25^o-25^o=130^o\)
Vậy ...
Vì \(\Delta ABC\) cân tại A có:
\(\widehat{ABC}=\widehat{BCA}=50^o\)
\(\rightarrow\widehat{BCA}=180^o-\left(\widehat{ABC}+\widehat{BCA}\right)\)
\(\Leftrightarrow\widehat{BAC}=180^o-\left(50+50\right)=80^o\)
Mà \(+\widehat{ABC}+\widehat{ABD}=180^0\) ( 2 góc kề bù )
\(\rightarrow50^o+\widehat{ABD}=180^o\)
\(\rightarrow\widehat{ABD}=180^o-50^o=130^o\)
\(+\widehat{ACB}+\widehat{ACE}=180^o\) (2 góc kề bù )
\(\rightarrow50^o+\widehat{ACE}=180^o\)
\(\rightarrow\widehat{ACE}=180^o-50^o=130^o\)
Vì \(BA=BD\rightarrow\Delta BAD\) cân tại B
\(\rightarrow\widehat{BDA}=\widehat{BAD}=\frac{\left(180^o-\widehat{ABD}\right)}{2}\)
\(\rightarrow\frac{180^o-130^o}{2}=25^o\)
Vì \(CA=CE\rightarrow\Delta CAE\) cân tại E
Mà \(\widehat{ACE}=\widehat{ABD}=130^o\)
\(\rightarrow\widehat{AEC}=\widehat{CAE}=25^o\)
Lại có : \(\widehat{DEA}=\widehat{DAB}+\widehat{BAC}+\widehat{CAE}\)
\(\Leftrightarrow25^o+80^o+25^o=130^o\)
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