∴ac=ac′,bc=b′c′

∠ACB =∠AC′B′= 90°

∫AC = AC’

∴∠ac′c=∠acc′

∠∠BCD+∠ACC ' = 180-∠ACB = 180-90 = 90

∠B′C′D+∠AC′C =∠AC′B′= 90

∠AC′C =∠ACC′

∴∠bcd=∠b′c′d

2. Make BH∨B' C' after B, and extend the line from C'D to H.

∴∠b′c′d=∠h=∠bcd

△ BCH is an isosceles triangle.

∴bc=bh=b′c′

∠B'C'D=∠H, ∠B'D'C'=∠BDH (equal vertex angles)

BH = B′C′

∴△b′c′d≌△bhd(aas)

∴bd=b′d