Giải thích các bước giải:
a.Ta có :$ AC\perp BD\to \widehat{AIB}=\widehat{BAD}=90^o$
$\to \Delta AIB\sim\Delta DAB(g.g)$
b.Vì ABCD là hình thang vuông tại A,B$\to AB//CD$
$\to \widehat{IAB}=\widehat{ICD},\widehat{IBA}=\widehat{IDC}$
$\to \Delta IAB\sim\Delta ICD(g.g)$
c.Ta có : $\widehat{ABD}=\widehat{DAC}(+\widehat{IAB}=90^o)$
$\widehat{BAD}=\widehat{ADC}=90^o$
$\to \Delta ABD\sim\Delta DAC(g.g)$
$\to \dfrac{AB}{AD}=\dfrac{AD}{DC}$
$\to AD^2=AB.DC=36\to AD=6$
$\to AC=\sqrt{AD^2+CD^2}=3\sqrt{13}$
Do $AB//CD\to \dfrac{IA}{IC}=\dfrac{AB}{CD}=\dfrac49$
$\to \dfrac{IA}{IA+IC}=\dfrac4{4+9}$
$\to \dfrac{IA}{AC}=\dfrac4{13}$
$\to IA=\dfrac4{13}AC=\dfrac{12\sqrt{13}}{13}$
$\to IC=AC-AI=\dfrac{27\sqrt{13}}{13}$
Vì $\Delta IAB\sim\Delta ICD$
$\to \dfrac{S_{IAB}}{S_{ICD}}=(\dfrac{AB}{CD})^2=\dfrac{16}{81}$
