Giải thích các bước giải:

a.Xét $\Delta FAD,\Delta FEA$ có:

Chung $\hat F$

$\widehat{FAE}=\widehat{FDA}(=90^o)$

$\to\Delta FAD\sim\Delta FEA(g.g)$

$\to\dfrac{FA}{FE}=\dfrac{FD}{FA}$

$\to AF\cdot AF=FD\cdot FE$

b.Xét $\Delta ABM,\Delta ADF$ có:

$\widehat{ABM}=\widehat{ADF}(=90^o)$

$AB=AD$

$\widehat{BAM}=90^o-\widehat{MAD}=\widehat{FAD}$

$\to\Delta ABM=\Delta ADF(g.c.g)$

$\to AM=AF$

Mà $AM\perp AF\to\Delta AMF$ vuông cân tại $A$

c.Ta có: $AD\perp EF, AE\perp AF$

$\to AD\cdot EF=AE\cdot AF=2S_{AEF}$

$\to (AD\cdot EF)^2=(AE\cdot AF)^2$

$\to  AD^2\cdot EF^2=AE^2\cdot AF^2$

$\to \dfrac1{EF^2}{AE^2\cdot AF^2}=\dfrac1{AD^2}$

$\to \dfrac1{AE^2+AF^2}{AE^2\cdot AF^2}=\dfrac1{AD^2}$ vì $EF^2=AE^2+AF^2$

$\to\dfrac1{AF^2}+\dfrac1{AE^2}=\dfrac1{AD^2}$

$\to\dfrac1{AF\cdot AF}+\dfrac1{AE\cdot AE}=\dfrac1{AD^2}$ không đổi khi $M$ di chuyển trên $BC$

d.Xét $\Delta FDA,\Delta FKC$ có:

Chung $\hat F$

$\widehat{FDA}=\widehat{FKC}(=90^o)$

$\to\Delta FDA\sim\Delta FKC(g.g)$

$\to\dfrac{FD}{FK}=\dfrac{FA}{FC}$

$\to\dfrac{FD}{FA}=\dfrac{FK}{FC}$

Mà $\widehat{KFD}=\widehat{AFC}$

$\to\Delta FDK\sim\Delta FAC(g.g)$

$\to \widehat{FKD}=\widehat{FCA}=\widehat{DCA}=45^o$