a)

Ta có:

$\dfrac{OC}{OA}=\dfrac{5}{3}$

$\dfrac{OB}{OD}=\dfrac{10}{6}=\dfrac{5}{3}$

$\to \dfrac{OC}{OA}=\dfrac{OB}{OD}=\dfrac{5}{3}$

Xét $\Delta OCB$ và $\Delta OAD$, ta có:

$\dfrac{OC}{OA}=\dfrac{OB}{OD}\,\,\,\left( cmt \right)$

$\widehat{BOD}$ là góc chung

$\to \Delta OCB\backsim\Delta OAD\,\,\,\left( c.g.c \right)$

b)

Vì $\Delta OCB\backsim\Delta OAD\,\,\,\left( cmt \right)$

$\to \widehat{OBC}=\widehat{ODA}$ ( hai góc tương ứng )

$\to \widehat{ABI}=\widehat{CDI}$

Xét $\Delta AIB$ và $\Delta CID$, ta có:

$\widehat{ABI}=\widehat{CDI}\,\,\,\left( cmt \right)$

$\widehat{AIB}=\widehat{CID}$ ( hai góc đối đỉnh )

$\to \Delta AIB\backsim\Delta CID\,\,\,\left( g.g \right)$

c)

Vì $\Delta AIB\backsim\Delta CID\,\,\,\left( cmt \right)$

$\to \dfrac{IA}{IC}=\dfrac{IB}{ID}$

$\to IA.ID=IB.IC$

d)

Vì $\Delta AIB\backsim\Delta CID$

$\to \dfrac{{{S}_{\Delta AIB}}}{{{S}_{\Delta CID}}}={{\left( \dfrac{AB}{CD} \right)}^{2}}={{\left( \dfrac{OB-OA}{OD-OC} \right)}^{2}}={{\left( \dfrac{10-3}{6-5} \right)}^{2}}=49$

$\to {{S}_{\Delta AIB}}=49{{S}_{\Delta CID}}=49.3=147\,\,\,\left( c{{m}^{2}} \right)$