Đáp án:
b)
Theo phần a) ta có $\widehat{BIC}=120^0$
mà $\widehat{BIK}+\widehat{BIC}=180^0$ (kề bù)
$\Rightarrow \widehat{BIK}=180-120=60^0$
Ta có:
$\widehat{B_{1}}+\widehat{B_2}+\widehat{B_3}+\widehat{B_4}=180^0$
mà $\widehat{B_{1}}=\widehat{B_2}$ (do $BD$ là phân giác)
$\widehat{B_3}=\widehat{B_4}$ (do $BK$ là phân giác)
$\Rightarrow 2\widehat{B_2}+2\widehat{B_3}=180^0$
$\Rightarrow \widehat{B_2}+\widehat{B_3}=90^0$
$\Rightarrow \widehat{IBK}=90^0$
Xét $\triangle BKI$ có:
$\widehat{IBK}+\widehat{BIK}+\widehat{BKI}=180^0$ (t/c tổng ba góc trong tam giác)
$\Rightarrow \widehat{BKI}=180^0-\widehat{IBK}-\widehat{BIK}=180^0-90^0-60^0=30^0$
