Angle DBC = angle DBA = x, angle ABC = 2x
AB = AC, so angle BCD = 2x
Angle ADB external to triangle BDC, so angle ADB = 3x
BD = BC, so angle BDC = 90 - x
angle ADB + angle BDC = 180 ==> 3x + (90 - x) = 180 ==> x = 45
==> A = 90 - x = \(\boxed{45}\)
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