Giải thích các bước giải:

a.Xét $\Delta AIB,\Delta AID$ có:

Chung $AI$

$\widehat{IAB}=\widehat{IAD}$

$AB=AC$

$\to \Delta AIB=\Delta AID(c.g.c)$

$\to IB=ID$

b.Từ a $\to \widehat{ABI}=\widehat{ADI}$

$\to \widehat{IBE}=180^o-\widehat{ABI}=180^o-\widehat{ADI}=\widehat{IDC}$

Xét $\Delta IBE,\Delta IDC$ có:

$\widehat{BIE}=\widehat{DIC}$

$IB=ID$

$\widehat{IBE}=\widehat{IDC}$

$\to \Delta IBE=\Delta IDC(g.c.g)$

c.Từ b $\to BE=CD\to AE=AB+BE=AD+DC=AC$

$\to \Delta ACE$ cân tại $A$

Mà $AB=AD\to \Delta ABD$ cân tại $A$

$\to \widehat{ABD}=90^o-\dfrac12\hat A=\widehat{AEC}$

$\to DB//CE$

d.Ta có:

$\widehat{ABC}=2\widehat{ACB}$

$\to \widehat{ADI}=2\widehat{ICD}$

$\to \widehat{ADI}-\widehat{ICD}=\widehat{ICD}$

$\to \widehat{DIC}=\widehat{DCI}$

$\to \Delta DIC$ cân tại $D$

$\to DI=DC$

$\to AB+BI=AD+ID=AD+DC=AC$