Giải thích các bước giải:

a.Ta có: $BA=BD\to \Delta BAD$ cân tại $B$

Mà $BI$ là phân giác $\hat B$

$\to BI\perp AD, BI$ là trung trực $AD$

$\to IA=ID, \widehat{BAD}=\widehat{BDA}$

Tương tự $\widehat{CAE}=\widehat{CEA}, IA=IE$

$\to \widehat{BAD}+\widehat{CAE}=(90^o-\dfrac12\hat B)+(90^o-\dfrac12\hat C)$ 

$\to \widehat{BAD}+\widehat{DAC}+\widehat{DAE}=180^o-\dfrac12(\hat B+\hat C)$

$\to \widehat{BAC}+\widehat{DAE}=180^o-\dfrac12\cdot 90^o$

$\to 90^o+\widehat{DAE}=135^o$

$\to \widehat{DAE}=45^o$

b.Từ a $\to IA=ID=IE$

$\to \widehat{DIE}=180^o-(\widehat{IDE}+\widehat{IED})$

$\to \widehat{DIE}=180^o-(\widehat{AED}-\widehat{IEA}+\widehat{ADE}-\widehat{IDA})$

$\to \widehat{DIE}=180^o-(\widehat{AED}-\widehat{IAE}+\widehat{ADE}-\widehat{IAD})$

$\to \widehat{DIE}=180^o-(\widehat{AED}+\widehat{ADE}-\widehat{IAD}-\widehat{IAE})$

$\to \widehat{DIE}=180^o-(\widehat{AED}+\widehat{ADE}-\widehat{DAE})$

$\to \widehat{DIE}=180^o-(\widehat{AED}+\widehat{ADE}+\widehat{DAE}-2\widehat{DAE})$

$\to \widehat{DIE}=180^o-(180^o-2\widehat{DAE})$

$\to \widehat{DIE}=2\widehat{DAE}=90^o$