Giải thích các bước giải:

c.Kẻ $BF\perp AI$

Xét $\Delta ABI,\Delta CBF$ có:

$\widehat{BAI}=\widehat{BCF}=90^o$

$BA=BC$

$\widehat{ABI}=90^o-\widehat{IBC}=\widehat{CBF}$

$\to\Delta ABI=\Delta CBF(g.c.g)$

$\to CF=AI, BI=BF$

Ta có $AI=3ID\to AI+ID=4DI\to 4ID=AD=a\to ID=\dfrac14a\to AI=\dfrac34a$

$\to CF=\dfrac34a$

Ta có $\widehat{ABI}=\widehat{IBK}, BE$ là phân giác $\widehat{CBK}$

$\to \widehat{IBE}=\widehat{IBK}+\widehat{KBE}=\dfrac12\widehat{KBA}+\dfrac12\widehat{KBC}=\dfrac12\widehat{ABC}=45^o=\dfrac12\widehat{IBF}$

$\to BE$ là phân giác $\widehat{IBF}$

Xét $\Delta BIE, \Delta FBE$ có:

Chung $BE$

$\widehat{IBE}=\widehat{EBF}$ vì $BE$ là phân giác $\widehat{IBF}$

$BI=BF$ 

$\to\Delta BIE=\Delta BFE(c.g.c)$

$\to FE=EI$

Đặt $CE=x, DE=a-x$

$\to IE=EF=CF+x=\dfrac34a+x$

Mà $IE^2=ID^2+DE^2$

$\to (\dfrac34a+x)^2=(\dfrac14a)^2+(a-x)^2$

$\to x=\dfrac{a}{7}$

$\to CE=\dfrac17a\to DE=\dfrac67a$

$\to S_{BIE}=S_{ABCD}-S_{ABI}-S_{DIE}-S_{BCE}$

$\to S_{BIE}=a^2-\dfrac12\cdot AB\cdot AI-\dfrac12\cdot DI\cdot DE-\dfrac12\cdot BC\cdot CE$

$\to S_{BIE}=a^2-\dfrac12\cdot a\cdot \dfrac34a-\dfrac12\cdot \dfrac14a\cdot \dfrac67a-\dfrac12\cdot a\cdot \dfrac17a$

$\to S_{BIE}=\dfrac{25}{56}a^2$