a) ABCD là hình chữ nhật nên \(\widehat{A}\)=90\(^0\)
=> \(\Delta ABD\)vuông tại A
Xét \(\Delta ABD\).Theo định lí pytago:
DB\(^2\)=AB\(^2\)+\(AD^2\)
= \(8^2\)+\(6^2\)
=100
=> DB=10(cm)
b) Xét \(\Delta ADH\)và \(\Delta ADB\) có:
\(\widehat{H2}\)=\(\widehat{BAH}\)(=90\(^0\))
\(\widehat{D}\)chung
=> \(\Delta HDA\sim\Delta ADB\)(g.g)
c) Theo câu c) \(\Delta HDA\sim\Delta ADB\)=> \(\dfrac{HD}{AD}\)=\(\dfrac{DA}{DB}\)=> AD.DA= HD.DB
=> AD\(^2\)=DH.DB
d)Xét \(\Delta AHB\)và \(\Delta BCD\) có:
\(\widehat{H1}\)=\(\widehat{BCD}\)(=90\(^0\))
\(\widehat{B1}\)=\(\widehat{D2}\)(so le trong, AB//CD, ABCD là hình chữ nhật)
=> \(\Delta AHB\)~\(\Delta BCD\)(g.g)
e) S\(_{ADB}\)= \(\dfrac{1}{2}\).AH.BD= \(\dfrac{1}{2}\)AB.AD
=> AH= \(\dfrac{AB.AD}{BD}\) => AH=\(\dfrac{8.6}{10}\)=4,8 (cm)
Theo câu b) \(\Delta HDA\sim\Delta ADB\)=>\(\dfrac{DH}{DB}\)=\(\dfrac{AH}{AB}\)
=> \(\dfrac{DH}{10}\)=\(\dfrac{4,8}{8}\)=> DH=69(cm)
