- Bổ đề: Cho \(\Delta ABC\) thi ta có: \(S_{ABC}=\dfrac{1}{2}AB.AC.\sin\widehat{A}\).
- C/m: Hạ đường cao BH.
- \(\Delta ABH\) vuông tại H có: \(\sin\widehat{A}=\dfrac{BH}{AB}\Rightarrow BH=AB.\sin\widehat{A}\).
- \(S_{ABC}=\dfrac{1}{2}BH.AC=\dfrac{1}{2}AB.AC.\sin\widehat{A}\)
- Áp dụng cho bài toán, ta có:
\(\left\{{}\begin{matrix}S_{ABD}=\dfrac{1}{2}AD.AB.\sin\widehat{\dfrac{A}{2}}\\S_{ACD}=\dfrac{1}{2}AD.AC.\sin\dfrac{\widehat{A}}{2}\\S_{ABC}=\dfrac{1}{2}AB.AC.\sin\widehat{A}\end{matrix}\right.\)
Mà \(S_{ABD}+S_{ACD}=S_{ABC}\)
\(\Rightarrow\dfrac{1}{2}.\left(AD.AB+AD.AC\right).\sin\dfrac{\widehat{A}}{2}=\dfrac{1}{2}AB.AC.\sin\widehat{A}\)
\(\Rightarrow\dfrac{\left(AD.AB+AD.AC\right)}{AB.AC}.\sin\dfrac{\widehat{A}}{2}=\sin\widehat{A}\)
\(\Rightarrow\left(\dfrac{AD}{AB}+\dfrac{AD}{AC}\right).\sin\dfrac{\widehat{A}}{2}=\sin\widehat{A}\left(đpcm\right)\)
