Đáp án:

a)

Ta có:

$\dfrac{AB}{BD}=\dfrac{AC}{CE}$

$\Rightarrow BC//DE$ (định lý Talet)

b)

Ta có:

$\widehat{MBD}=\widehat{ABC}$

$\widehat{NCE}=\widehat{ACB}$

mà $\widehat{ABC}=\widehat{ACB}$

$\Rightarrow \widehat{MBD}=\widehat{NCE}$

Xét $\triangle BMD$ và $\triangle CNE$ có

$\widehat{BMD}=\widehat{CNE}=90^0$

$BD=CE$

$\widehat{MBD}=\widehat{NCE}$

$\Rightarrow \triangle BMD=\triangle CNE$ (cạnh huyền - góc nhọn)

$\Rightarrow BM=CN$

c)

Ta có:

$\widehat{ABM}+\widehat{ABC}=180^0$

$\widehat{ACN}+\widehat{ACB}=180^0$

mà $\widehat{ABC}=\widehat{ACB}$

$\Rightarrow \widehat{ABM}=\widehat{ACN}$

Xét $\triangle ABM$ và $\triangle ACN$ có

$AB=AC$

$\widehat{ABM}=\widehat{ACN}$

$BM=CN$

$\Rightarrow \triangle ABM=\triangle ACN$ (c.g.c)

$\Rightarrow AM=AN$

$\Rightarrow \triangle AMN$ cân tại $A$.