a) Xét \(\Delta AHB\) và \(\Delta BCD\) có:
\(\widehat{AHB}=\widehat{BCD}=90^0\)
\(\widehat{ABH}=\widehat{BDC}\) (cùng phụ với góc DBC )
suy ra: \(\Delta AHB~\Delta BCD\)
b) Xét \(\Delta ADB\) và \(\Delta HDA\) có:
\(\widehat{DAB}=\widehat{DHA}=90^0\)
\(\widehat{ADB}\) CHUNG
suy ra: \(\Delta ADB~\Delta HDA\)
\(\Rightarrow\)\(\frac{AD}{HD}=\frac{DB}{DA}\)
\(\Rightarrow\)\(AD^2=DH.DB\) (ĐPCM)
c) Áp dụng định lý Pytago ta có:
\(BD^2=AD^2+AB^2\)
\(\Leftrightarrow\)\(BD^2=6^2+8^2=100\)
\(\Leftrightarrow\)\(BD=\sqrt{100}=10\) cm
\(\Delta ADB~\Delta HDA\) \(\Rightarrow\) \(\frac{AD}{HD}=\frac{AB}{HA}=\frac{DB}{DA}\)
hay \(\frac{6}{HD}=\frac{8}{HA}=\frac{10}{6}=\frac{5}{3}\)
suy ra: \(DH=3.6cm\) \(AH=4,8cm\)
