a: ΔABC vuông tại A
=>\(\widehat{ABC}+\widehat{ACB}=90^0\)
=>\(\widehat{ACB}=30^0\)
Xét ΔABC vuông tại A có \(sinC=\dfrac{AB}{BC}\)
=>\(\dfrac{4}{BC}=sin30=\dfrac{1}{2}\)
=>BC=8(cm)
\(AC=\sqrt{BC^2-AB^2}=4\sqrt{3}\left(cm\right)\)
Xét ΔABC vuông tại A có AH là đường cao
nên AH*BC=AB*AC
=>\(AH\cdot8=4\cdot4\sqrt{3}=16\sqrt{3}\)
=>\(AH=2\sqrt{3}\left(cm\right)\)
Xét ΔABC vuông tại A có AH là đường cao
nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot CB\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=\dfrac{4^2}{8}=2\left(cm\right)\\CH=\dfrac{48}{8}=6\left(cm\right)\end{matrix}\right.\)
b: \(BC\cdot sinB\cdot sinC\)
\(=BC\cdot\dfrac{AC}{BC}\cdot\dfrac{AB}{BC}=\dfrac{AB\cdot AC}{BC}=AH\)
\(BC\cdot cos^2B\)
\(=BC\cdot\left(\dfrac{AB}{BC}\right)^2=\dfrac{AB^2}{BC}=BH\)
\(BC\cdot sin^2B=BC\cdot\left(\dfrac{AC}{BC}\right)^2=\dfrac{AC^2}{BC}=CH\)
c:
\(\dfrac{AH^2}{AC^2}=\dfrac{HB\cdot HC}{BC\cdot HC}=\dfrac{HB}{BC}\)
ΔHAB vuông tại H có HD là đường cao
nên \(\left\{{}\begin{matrix}BD\cdot BA=BH^2\\AD\cdot AB=AH^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BD=\dfrac{BH^2}{AB}\\AD=\dfrac{AH^2}{AB}\end{matrix}\right.\)
ΔHAC vuông tại H có HE là đường cao
nên \(\left\{{}\begin{matrix}CE\cdot CA=CH^2\\AE\cdot AC=AH^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}CE=\dfrac{CH^2}{AC}\\AE=\dfrac{AH^2}{AC}\end{matrix}\right.\)
\(\dfrac{DB}{EC}=\dfrac{HB^2}{AB}:\dfrac{HC^2}{AC}\)
\(=\dfrac{HB^2}{AB}\cdot\dfrac{AC}{HC^2}\)
\(=\left(\dfrac{HB}{HC}\right)^2\cdot\dfrac{AC}{AB}=\dfrac{AC}{AB}\cdot\left(\dfrac{AB}{AC}\right)^4=\left(\dfrac{AB}{AC}\right)^3\)
\(BD\cdot CE\cdot BC\)
\(=\dfrac{BH^2}{AB}\cdot\dfrac{CH^2}{AC}\cdot BC\)
\(=\dfrac{AH^4}{AH}=AH^3\)
=DE3
\(BC\cdot HD\cdot HE\)
\(=BC\cdot\dfrac{HA\cdot HB}{AB}\cdot\dfrac{HA\cdot HC}{AC}\)
\(=\dfrac{1}{AH}\cdot\dfrac{HA^2\cdot HB\cdot HC}{1}=\dfrac{HA\cdot HB\cdot HC}{1}=HA^3\)
\(=DE^3\)
=>ĐPCM
