Giải thích các bước giải:
a.Xét $\Delta AIB,\Delta DAB$ có:
$\widehat{AIB}=\widehat{BAD}(=90^o)$
Chung $\hat B$
$\to\Delta AIB\sim\Delta DAB(g.g)$
b.Xét $\Delta IAB,\Delta ICD$ có:
$\widehat{AIB}=\widehat{DIC}$
$\widehat{IBA}=\widehat{IDC}$ vì $AB//CD$
$\to\Delta IAB\sim\Delta ICD(g.g)$
c.Xét $\Delta ABD,\Delta ADC$ có:
$\widehat{DAB}=\widehat{ADC}(=90^o)$
$\widehat{ADB}=90^o-\widehat{IDC}=\widehat{ICD}=\widehat{ACD}$
$\to\Delta ABD\sim\Delta DAC(g.g)$
$\to\dfrac{AB}{DA}=\dfrac{AD}{CD}$
$\to AD^2=AB\cdot CD$
d.Ta có:
$AD^2=AB\cdot CD\to AD^2=36\to AD=6$
$\to BD=\sqrt{AB^2+AD^2}=2\sqrt{13}$
Từ câu a $\to\dfrac{AB}{DB}=\dfrac{IB}{AB}$
$\to IB=\dfrac{AB^2}{BD}=\dfrac8{\sqrt{13}}$
$\to IA=\sqrt{AB^2-IB^2}=\dfrac{12}{\sqrt{13}}$
Ta có:
$AC=\sqrt{AD^2+DC^2}=3\sqrt{13}$
$\to IC=AC-AI=\dfrac{27}{\sqrt{13}}$
Từ câu b
$\to\dfrac{S_{IAB}}{S_{ICD}}=(\dfrac{AB}{CD})^2=\dfrac{16}{81}$
