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 $\Delta AIB\backsim\Delta CID$
$\to \dfrac{{{S}_{\Delta AIB}}}{{{S}_{\Delta CID}}}={{\left( \dfrac{AB}{CD} \right)}^{2}}={{\left( \dfrac{OB-OA}{OD-OC} \right)}^{2}}={{\left( \dfrac{10-3}{6-5} \right)}^{2}}=49$
$\to {{S}_{\Delta AIB}}=49{{S}_{\Delta CID}}=49.3=147\,\,\,\left( c{{m}^{2}} \right)$
