cho đường tròn (O) đường kính AB .Kẻ dây CD vuông góc với AB không đi qua tâm sao cho góc COD=1400 .Từ điểm C và D kẻ dây CE và DF lần lượt song song với AB hãy tính các góc ADE : CAD :biết rằng góc EAF=700
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Kẻ phân giác AD, BK vuông góc với AD.
\(\sin\frac{\widehat{A}}{2}=\sin BAD\)
Xét tam giác AKB vuông tại K, ta có:
\(\sin BAD=\frac{BK}{AK}\left(1\right)\)
Xét tam giác BKD vuông tại K, ta có:
\(BK\Leftarrow BD\)thay vào (1)
\(\sin BAD\Leftarrow\frac{BD}{AB}\left(2\right)\)
Lại có: \(\frac{BD}{CD}=\frac{AB}{AC}\)
\(\Rightarrow\frac{BD}{\left(BD+CD\right)}=\frac{AB}{\left(AB+AC\right)}\)
\(\Rightarrow\frac{BD}{BC}=\frac{AB}{\left(AB+AC\right)}\)
\(\Rightarrow BD=\frac{\left(AB.BC\right)}{\left(AB+AC\right)}\)thay vào (2)
\(\sin BAD\Leftarrow\frac{\left[\frac{\left(AB.BC\right)}{\left(AB+AC\right)}\right]}{AB}\)
\(=\frac{BC}{\left(AB+AC\right)}\left(ĐPCM\right)\)
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