Kẻ tia phân giác BK cắt AC tại K
\(\Rightarrow\widehat{ABK}=\widehat{CBK}=\dfrac{\widehat{ABC}}{2}\)
Mà ta có \(\widehat{B}=2\widehat{C}\)
Suy ra \(\widehat{ABK}=\widehat{KBC}=\widehat{KCB}\)
Xét △BKC có
\(\widehat{KBC}=\widehat{KCB}\)(cmt)
Suy ra △BKC cân tại K\(\Rightarrow BK=KC\)
Xét △ABK và △ACB có
\(\widehat{A}\) chung
\(\widehat{ABK}=\widehat{KCB}\)(cmt)
Suy ra △ABK ∼ △ACB(g.g)
\(\Rightarrow\dfrac{AB}{AC}=\dfrac{BK}{BC}\Rightarrow AC.BK=AB.BC\Rightarrow AC.BK=8.10=80\Rightarrow AC.KC=80\left(1\right)\)
Ta có △ABK ∼ △ACB\(\Rightarrow\dfrac{AB}{AC}=\dfrac{AK}{AB}\Rightarrow AC.AK=AB^2\Rightarrow AC.AK=8^2=64\left(2\right)\)
Cộng (1),(2)\(\Rightarrow AC.KC+AC.AK=80+64\Rightarrow AC\left(KC.AK\right)=144\Rightarrow AC.AC=144\Rightarrow AC^2=144\Rightarrow AC=12\left(cm\right)\)
b) Giả sử AC>BC>AB
Đặt AB=x(x∈N*)\(\Rightarrow BC=x+1\Rightarrow AC=x+2\)
Theo câu a, ta có △ABK ∼ △ACB
\(\Rightarrow\dfrac{AB}{AC}=\dfrac{BK}{BC}\Rightarrow AB.BC=BK.AC\Rightarrow AB.BC=KC.AC\Rightarrow x\left(x+1\right)=\left(x+2\right)KC\left(3\right)\)
ta có △ABK ∼ △ACB\(\Rightarrow\dfrac{AB}{AC}=\dfrac{AK}{AB}\Rightarrow AB^2=AK.AC\Rightarrow x^2=\left(x+2\right)AK\left(4\right)\)
Cộng (3),(4)\(\Rightarrow x\left(x+1\right)+x^2=\left(x+2\right)KC+\left(x+2\right).AK\Leftrightarrow x^2+x+x^2=\left(x+2\right)\left(KC+AK\right)\Leftrightarrow2x^2+x=\left(x+2\right).AC\Leftrightarrow2x^2+x=\left(x+2\right)^2\Leftrightarrow2x^2+x=x^2+4x+4\Leftrightarrow x^2-3x-4=0\Leftrightarrow x^2+x-4x-4=0\Leftrightarrow x\left(x+1\right)-4\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x-4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Vậy x=4\(\Rightarrow AB=4\Rightarrow BC=5\Rightarrow AC=6\)
