Giải thích các bước giải:
$a)ABCD$ là hình chữ nhật
$\Rightarrow CD=AB=12cm;AD=BC=9cm;AB//CD\\ \Rightarrow \widehat{B_1}=\widehat{D_1}$
Xét $\Delta AHB$ và $\Delta BCD$
$\widehat{B_1}=\widehat{D_1}\\ \widehat{AHB}=\widehat{BCD}=90^\circ\\ \Rightarrow \Delta AHB \backsim \Delta BCD\\ b)\Delta AHB \backsim \Delta BCD\\ \Rightarrow \dfrac{AH}{BC}=\dfrac{AB}{BD}\\ \Rightarrow AH=\dfrac{AB.BC}{BD}$
$\Delta ABD$ vuông tại $A$
$\Rightarrow BD=\sqrt{AB^2+AD^2}=15cm\\ \Rightarrow AH=\dfrac{AB.BC}{BD}=7,2cm\\ c)\widehat{B_1}=\widehat{A_2}=90^\circ\\ \widehat{A_1}=\widehat{A_2}=90^\circ\\ \Rightarrow \widehat{B_1}=\widehat{A_1}$
Xét $\Delta AHD$ và $\Delta BHA$
$\widehat{H_1}=\widehat{H_2}=90^\circ\\ \widehat{B_1}=\widehat{A_1}\\ \Rightarrow \Delta AHD \backsim \Delta BHA\\ \Rightarrow \dfrac{AH}{BH}=\dfrac{HD}{HA}\\ \Rightarrow AH^2=HB.HD\\ d)S_{\Delta ABD}=\dfrac{1}{2}BD.AH\\ S_{\Delta ABD}=\dfrac{1}{2}AD.AB\\ \Rightarrow BD.AH=AD.AB$
