Because CD=C'D', AB = a' b' and AD = a' d'
In Rt△CED and Rt△C'E'D', CE = C 'e' and AD = A'd', so Rt△CED≌Rt△C'E'D'(HL).
So ∠CAD=∠C'A'D', CD = C'd' and AD = A'd', so △ CDA △ C'd 'a (SAS).
So AC=A'C', in Rt△ABC and Rt△A'B'C, AC = a' c, AB = a' b'
So Rt△ABC≌Rt△A'B'C'(HL)
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