BD=BC=AD
= & gt∠A=∠ABD，∠BDC=∠C=(180 -∠A)/2
= & gt∠DBC=90 -(3/2)∠A
∠DBC+∠C+∠BDC=180
= & gt∠A=36
2
AB=AC，AE=AF
= & gtAB-AE=AC-AF，∠B=∠C
即EB=FC，
d是BC的中點= & gtBD=DC
= & gt△EDB?△FDC
∠EDB=∠FDC
三
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