∠ BAC =∠ B = 67.5, ∠ BAC' = 22.5 After flipping, ∴AC'⊥BD.

∴∠C'PD=90 ∵DC=2,∴PD=PC'=√2

∵AC' is both an angular bisector of △ABD and a vertical line, so △ABD is an isosceles triangle, ∴BP=PD=√2.

∴BD=2√2

(2) The same idea as above, transfer △ACD to △ AC' D and hand over BD to P.

It can be deduced that ∠ BDC' = 90, PD = 2 ∠ 3/3,

AC=2√3+2 can be derived from DC=2, and the length of AD can be derived from the value of PD.

Then according to △BAD∽△BCA, other edges can be deduced (I won't write this in detail, it's too troublesome).