Giải thích các bước giải:

Ta có: $\Delta ABC$ vuông tại $B, M$ là trung điểm $BC$

$\to MB=MA=MC=\dfrac12AC$

$\to \Delta MBC$ cân tại $M$

$\to \widehat{NBI}=\widehat{MBC}=\hat C=90^o-\widehat{HBC}=\widehat{HBK}=\widehat{IKB}$

$\to\Delta INB\sim\Delta IBK(g.g)$

$\to \widehat{INB}=\widehat{IBK}=90^o$

$\to MB\perp IK$

Ta có:

$AC=\sqrt{AB^2+BC^2}=25$

$BH.AC=BA.BC(=2S_{ABC})\to BH=\dfrac{AB.BC}{AC}=12$

$AH=\sqrt{AB^2-HB^2}=9$

$HK.AB=HB.HA(=2S_{HAB})\to HK=\dfrac{HA.HB}{AB}= 7.2$

$\to IB=HK=7.2$

$\to KB=\sqrt{HB^2-HK^2}=\sqrt{12^2-7.2^2}=9.6$

$KI=HB=12$

$NB=\dfrac{BK.BI}{IK}=\dfrac{9.6\cdot 7.2}{12}=5.76$

$KN=\sqrt{KB^2-NB^2}=\sqrt{9.6^2-5.76^2}=7.68$

$\to S_{BKN}=\dfrac12\cdot 5.76\cdot 7.68=22.1184$