Xét $ΔABC$ có:
$\widehat{ABC} + \widehat{BCA} + \widehat{CAB} = 180^\circ$
$\to \widehat{CAB} = 180^\circ - (\widehat{ABC} + \widehat{BCA}) = 180^\circ - (75^\circ + 45^\circ) = 60^\circ$
Xét $ΔABD$ vuông tại $D$ có:
$\widehat{DAB} = \widehat{CAB} = 60^\circ$
$\to \widehat{ABD} = 90^\circ - \widehat{DAB} =90^\circ - 60^\circ = 30^\circ$
Ta lại có:
$\widehat{ACE} = \widehat{ABD}$ (cùng phụ $\widehat{ABC}$)
$\to \widehat{ACE} =30^\circ$
Do đó:
$\widehat{IB} + \widehat{ICB}$
$=(\widehat{ABC} - \widehat{ABD}) + (\widehat{BCA} - \widehat{ACE})$
$=(75^\circ - 30^\circ) + (45^\circ - 30^\circ)$
$= 45^\circ + 15^\circ$
$=60^\circ$
