A) HC=?
Áp dụng định lý Py-ta-go vào \(\Delta\)HBC vuông ở H:
=>BH2+HC2=BC2
=>122+HC2=152
=>HC2=152-122=81
=>HC=9
B)\(\dfrac{DH}{BH}=\dfrac{16}{12}=\dfrac{4}{3};\dfrac{BH}{HC}=\dfrac{12}{9}=\dfrac{4}{3}\)
\(\Rightarrow\dfrac{DH}{BH}=\dfrac{BH}{HC}\)
Xét \(\Delta HDB\) và\(\Delta HBC\) đều vuông ở H:
\(\dfrac{DH}{BH}=\dfrac{BH}{HC}\left(cmt\right)\)
=>\(\Delta HDB\)\(\sim\)\(\Delta HBC\)
=>\(\widehat{DBH}=\widehat{BCH}\)
Ta có:\(\widehat{BCH}+\widehat{CBH}=90^0\)(\(\Delta HBC\)vuông)
=>\(\widehat{DBH}+\widehat{CBH}=90^0\)
=>\(DB\perp BC\)