4BA=3BC
=>BA/BC=3/4
Xét ΔABC vuông tại B có BH là đường cao
nên BA^2=AH*AC và BC^2=CH*CA
=>AH/CH=(BA/BC)^2=9/16
=>AH/9=CH/16=k
=>AH=9k; CH=16k
ΔBAC vuông tại B có BH là đường cao
nên BH^2=HA*HC
=>144k^2=(12/5)^2=144/25
=>k^2=1/25
=>k=1/5
=>AH=1,8cm; CH=3,2cm
AC=1,8+3,2=5cm
\(BA=\sqrt{AH\cdot AC}=\sqrt{1.8\cdot5}=3\left(cm\right)\)
\(BC=\sqrt{3.2\cdot5}=4\left(cm\right)\)
\(4AB=3BC\Rightarrow AB=\dfrac{3}{4}BC\)
\(\dfrac{1}{BH^2}=\dfrac{1}{AB^2}+\dfrac{1}{BC^2}\)
\(\Leftrightarrow\dfrac{1}{BH^2}=\dfrac{AB^2+BC^2}{AB^2.BC^2}\)
\(\Leftrightarrow BH^2=\dfrac{AB^2.BC^2}{AB^2+BC^2}\)
\(\Leftrightarrow BH^2=\dfrac{\dfrac{9}{16}BC^2.BC^2}{\dfrac{9}{16}BC^2+BC^2}\)
\(\Leftrightarrow BH^2=\dfrac{\dfrac{9}{16}BC^4}{\dfrac{25BC^2}{16}}=\dfrac{9}{25}BC^2\)
\(\Leftrightarrow BC^2=\dfrac{25BH^2}{9}=\dfrac{25.\dfrac{144}{25}}{9}=16\)
\(\Rightarrow BC=4\left(cm\right)\)
\(\Rightarrow AB=\dfrac{3}{4}BC=\dfrac{3}{4}.4=3\left(cm\right)\)
\(AC^2=AB^2+BC^2=9+16=25\left(cm\right)\)
\(\Rightarrow AC=5\left(cm\right)\)
\(AB^2=AC.AH\Rightarrow AH=\dfrac{AB^2}{AC}=\dfrac{9}{5}\left(cm\right)\)
\(\Rightarrow CH=AC-AH=5-\dfrac{9}{5}=\dfrac{16}{5}\left(cm\right)\)
