Giải thích các bước giải:
a.Ta có :
$MA=MN, MB=MC\to\dfrac{MD}{MN}=\dfrac{MB}{MC}\to AB//CN\to\dfrac{AB}{CN}=\dfrac{AM}{MN}=1\to AB=CN$
$\to\widehat{ACN}+\widehat{BAC}=180^o$
Lại có $\widehat{BAC}+\widehat{AEC}=\widehat{BAC}+\widehat{DAC}=90^o$
$\to\widehat{BAC}+\widehat{CAE}+\widehat{DAB}+\widehat{BAC}=180^o\to\widehat{DAE}+\widehat{BAC}=180^o$
$\to\widehat{BAC}+\widehat{DAC}=\widehat{DAE}+\widehat{BAC}\to\widehat{DAE}=\widehat{ACN}$
Mà $AD=AC, AE=AB=CN\to\Delta DAE=\Delta ACN(c.g.c) $
b.Thiếu dữ kiện để trả lời
c.Theo câu a $\to\widehat{CAN}=\widehat{ADE}\to \widehat{ADI}+\widehat{DAI}=\widehat{DAI}+\widehat{IAP}=90^o$
$\to AM\perp DE\to AB^2-IE^2=AE^2-IE^2=AI^2=AD^2-DI^2\to AB^2+DI^2=IE^2+AE^2$