a/ Xét \(ΔHBA\) và \(ΔABC\):
\(\widehat B:chung\)
\(\widehat{BHA}=\widehat{BAC}(=90^\circ)\)
\(→ΔHBA\backsim ΔABC(g-g)\)
\(→\dfrac{AB}{BH}=\dfrac{BC}{AB}\)
\(↔AB^2=BH.BC\)
b/ Áp đụng định lý Pytago vào \(ΔABC\) vuông tại \(A\)
\(→BC=\sqrt{AB^2+AC^2}=\sqrt{12^2+16^2}=\sqrt{400}=20(cm)\)
\(AD\) là đường phân giác \(\widehat A\)
\(→\dfrac{AD}{CD}=\dfrac{BA}{BC}\) hay \(\dfrac{AD}{CD}=\dfrac{12}{20}=\dfrac{3}{5}\)
\(→\dfrac{AD}{3}=\dfrac{CD}{5}=\dfrac{AD+CD}{3+5}=\dfrac{AC}{8}=\dfrac{16}{8}=2\)
\(→\begin{cases}AD=6(cm)\\CD=10(cm)\end{cases}\)
c/ \(ΔHBA\backsim ΔABC\)
\(→\dfrac{BA}{BC}=\dfrac{BH}{BA}\)
Xét \(ΔBHE\) và \(ΔBAD\):
\(\widehat{HBE}=\widehat{ABD}\) (\(BD\) hay \(BE\) là đường phân giác \(\widehat B\) )
\(\widehat{BHE}=\widehat{BAD}(=90^\circ)\)
\(→ΔBHE\backsim ΔBAD(g-g)\)
\(→\dfrac{BH}{BE}=\dfrac{BA}{BD}\)
\(→\dfrac{BH}{BA}=\dfrac{BE}{BD}\)
mà \(\dfrac{BH}{BA}=\dfrac{BA}{BC}\)
\(→\dfrac{BA}{BC}=\dfrac{BE}{BD}\)
mà \(\dfrac{BA}{BC}=\dfrac{DA}{DC}\)
\(→\dfrac{DA}{DC}=\dfrac{BE}{BD}\)
\(↔DA.BD=DC.BE\)
