Đề đúng là: \(\dfrac{AH}{AC}=\dfrac{3}{5}\). Bạn tự vẽ hình nhé.
(a) Theo đề: \(\dfrac{AH}{AC}=\dfrac{3}{5}\Leftrightarrow AC=\dfrac{5}{3}AH\)
Ta có: \(BC^2=AB^2+AC^2\left(Pythagoras\right)\)
\(\Leftrightarrow BC^2=15^2+\left(\dfrac{5}{3}AH\right)^2\Rightarrow BC=\sqrt{225+\dfrac{25}{9}AH^2}\)
Lại có: \(AB^2=BC.HB\Leftrightarrow HB=\dfrac{AB^2}{BC}=\dfrac{15^2}{\sqrt{225+\dfrac{25}{9}AH^2}}\)
Ta cũng có: \(AH^2=HB.HC=HB\left(BC-HB\right)=BC.HB-HB^2\)
\(\Leftrightarrow AH^2=\sqrt{225+\dfrac{25}{9}AH^2}\cdot\dfrac{15^2}{\sqrt{225+\dfrac{25}{9}AH^2}}-\left(\dfrac{15^2}{\sqrt{225+\dfrac{25}{9}AH^2}}\right)^2\)
\(=15^2-\dfrac{15^4}{225+\dfrac{25}{9}AH^2}\)
\(\Rightarrow AH=12\left(cm\right)\)
Thay vào tính được: \(HB=9\left(cm\right);BC=25\left(cm\right)\)
\(\Rightarrow HC=BC-HB=25-9=16\left(cm\right)\)
(b) Xét \(\Delta AHB\) vuông tại \(H:BE.AB=HB^2\Leftrightarrow BE=\dfrac{HB^2}{AB}\)
Tương tự, \(\Delta AHC\) vuông tại \(H:CF.AC=HC^2\Leftrightarrow CF=\dfrac{HC^2}{AC}\)
Ta có: \(BC.BE.CF=\left(\dfrac{AB.AC}{AH}\right)\cdot\dfrac{HB^2}{AB}\cdot\dfrac{HC^2}{AC}\)
\(=\dfrac{HB^2.HC^2}{AH}=\dfrac{\left(HB.HC\right)^2}{AH}=\dfrac{\left(AH^2\right)^2}{AH}=AH^3\left(đpcm\right)\)
