∴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