Đáp án:
+ Ta có:
$AB=AE+BE$
$AC=AD+DC$
mà $AE=AD;AB=AC$
$\Rightarrow BE=DC$
+ $\triangle EBC=\triangle DCB$ (c.g.c)
$\Rightarrow \widehat{ECB}=\widehat{DBC}$
$\Rightarrow \triangle IBC$ cân tại $I$
$\Rightarrow IB=IC$
+ $\triangle ABI=\triangle ACI$ (c.c.c)
$\Rightarrow \widehat{EAI}=\widehat{CAI}$
+ $\triangle EAI=\triangle DAI$ (c.g.c)
$\Rightarrow \widehat{EIA}=\widehat{AID}$ (1)
+ $\triangle IBM=\triangle ICM$ (c.c.c)
$\Rightarrow \widehat{BIM}=\widehat{CIM}$(2)
+ $\widehat{EID}=\widehat{BIC}$ (3)
Từ (1), (2), (3) suy ra $\widehat{EIA}=\widehat{CIM}$
mà $\widehat{EIA}+\widehat{AIC}=180^0$
$\Rightarrow \widehat{CIM}+\widehat{AIC}=180^0$
$\Rightarrow \widehat{AIM}=180^0$
hay $A,I,M$ thẳng hàng
