∫ In square ∠ABCD, BD divides∠ ABC equally.

∴PG=PH

∠∠PGB =∠ABC =∠PHB

So in the rectangular BGPH, ∠GPH=90=∠EPF means ∠EPG=∠FPH.

Then it can be proved that △EPG and △FPH are congruent. The second problem is solved by Pythagoras in an isosceles right triangle, and the answer is √2(BF+BE)/2=BP.

P.S. If you don't understand, you can ask _