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

That is, EB=FC,

D is the midpoint of BC = & gtBD=DC.

= & gt△EDB?△FDC

∠EDB=∠FDC

three

/question/60573470.html？ fr=qrl