Giải thích các bước giải:
Ta có: $AH\perp BC, HA=HD\to \Delta AHD$ vuông cân tại $H$
$\to \widehat{ADH}=45^o\to \widehat{ADC}=180^o-\widehat{ADH}=135^o$
Ta có: $\widehat{CDE}=90^o=\widehat{CAB},\widehat{ECD}=\widehat{ACB}$
$\to \Delta CDE\sim\Delta CAB(g.g)$
$\to \dfrac{CD}{CA}=\dfrac{CE}{CB}\to \dfrac{CD}{CE}=\dfrac{CA}{CB}$
Mà $\widehat{ACD}=\widehat{ECB}$
$\to \Delta CDA\sim\Delta CEB(c.g.c)$
$\to \widehat{BEC}=\widehat{ADC}=135^o$
$\to \widehat{AEB}=180^o-\widehat{BEC}=45^o$
Lai có: $\Delta ABE$ vuông tại $A\to \Delta ABE$ vuông cân
$\to AB=AE$
Vì $M$ là trung điểm $BE\to AM\perp BE$
$\to \widehat{AMF}=\widehat{FHB}(=90^o)$
Mà :$ \widehat{AFM}=\widehat{BFH}$
$\to \Delta AFM\sim\Delta BFH(g.g)$
$\to \dfrac{FA}{FB}=\dfrac{FM}{FH}$
$\to \dfrac{FA}{FM}=\dfrac{FB}{FH}$
Lại có: $\widehat{AFB}=\widehat{MFH}$
$\to \Delta AFB\sim\Delta MFH(c.g.c)$
$\to \widehat{FHM}=\widehat{ABF}$
$\to \widehat{AHM}=\widehat{ABE}=45^o$
