Certificate: Take a little E from BC line, make BE=AD, and then connect with DE.

From ∠ BAC = 90, AB=AC=BD, ∠ Abd = 30.

∠CAD=∠DBE= 15

BE=AD，BD=AC。

So △DBE is equal to△ △CAD.

So ∠ACD=∠BDE, DC=DE, that is ∠ DCE = ∠ DEC = ∠ DBE+∠ BDE =15+∠ ACD.

45 =∠ACB =∠DCE+∠ACD = 15+2∠ACD。

So ∠ ACD = 15 = ∠ DAC.

That is, AD=CD