Xét tứ giác $ABCD$:
$\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^\circ$
$→\widehat{A}+\widehat{B}+80^\circ+70^\circ=360^\circ$
$→\widehat{A}+\widehat{B}=210^\circ$
$AI;BI$ là phân giác $\widehat{A},\widehat{B}$
$→\widehat{IAB}+\widehat{IBA}=\dfrac{\widehat{A}+\widehat{B}}{2}=\dfrac{210^\circ}{2}=105^\circ$
Xét $ΔAIB$:
$\widehat{AIB}+\widehat{IAB}+\widehat{IBA}=180^\circ$
$→\widehat{AIB}+105^\circ=180^\circ$
$→\widehat{AIB}=180^\circ-105^\circ=75^\circ$
Vậy $\widehat{AIB}=75^\circ$
