1, ① The option is correct.
∠∠ea b+∠BAP = 90,∠BAP+∠PAD=90 ,∴∠EAB=∠PAD
* AE = AP,AB=AD,∴△APD≌△AEB.
2.② Wrong option.
The distance from b to AE is the distance from d to AP;
∵△APD?△aeb, ∴∠ABE=∠ADE,∴∠BED=∠BAD, that is, EB⊥ED (option ③ authentication).
∵EP=√2,BP=√5,∴EB=√3=PD
S _ (△ APD) =1/2×√ 3×√ 2/2 = √ 6/4 =1/2× AP× h, and the distance from ∴D to AP = √ 6/2.
∴ The distance from B to AE is also ∴ 6/2;
3. The correctness of option ③ has been proved in the previous proof;
4.④ Wrong option
s_(△apd)=√6/4,s_(△ape)= 1/2,∴s_(△apd)+s_(△ape)=√6/4+ 1/2;
5. The options are correct.
∵BE=√3,ED=√2+√3,∴BD=√(8+2√6)? ,
∴S_(□ABCD)=4+√6
(If you can't see clearly, the data in the picture has a standard format. )