设∠ABD=X
∵EB=ED,∠ABD=X
∴∠EDB=∠ABD=X
∴∠AED=∠ABD+∠EDB=2X
∵AD=ED
∴∠A=∠AED=2X
∴∠ADE=180-(∠A+∠AED)=180-4X
∵AB=AC
∴∠C=∠ACB=(180-∠A)/2=90-X
∵BC=BD
∴∠BDC=∠C=90-X
∵∠ADE+∠EDB+∠BDC=180
∴180-4X+X+90-X=180
∴X=45/2
∴∠A=2X=45
设∠ABD=X
∵EB=ED,∠ABD=X
∴∠EDB=∠ABD=X
∴∠AED=∠ABD+∠EDB=2X
∵AD=ED
∴∠A=∠AED=2X
∴∠ADE=180-(∠A+∠AED)=180-4X
∵AB=AC
∴∠C=∠ACB=(180-∠A)/2=90-X
∵BC=BD
∴∠BDC=∠C=90-X
∵∠ADE+∠EDB+∠BDC=180
∴180-4X+X+90-X=180
∴X=45/2
∴∠A=2X=45